Claim: $f$ is decreasing Let $a>b$. By taking $x=a$, there is some $c$ with $af(c)+cf(a)\le2$. By taking $x=c$, we know that $cf(b)+bf(c)>2$ since $a\ne b$. Then $2<cf(b)+af(c)$, so by adding this to $af(c)+cf(a)\le2$, we get $f(a)<f(b)$. Now we can claim that $\boxed{f(x)=\frac1x}$, which is indeed a solution (by AM-GM, we have $xf(y)+yf(x)\ge2\sqrt{xf(x)yf(x)}=2$ with equality iff $x=y$). Let $u\in\mathbb R^+$ be arbitrary, proving $f(u)=\frac1u$ suffices. Suppose $f(u)>\frac1u$. Then let $2\ge uf(v)+vf(u)$ by taking $x=u$. We have: $$2\ge uf(v)+vf(u)>uf(v)+\frac vu\ge2\sqrt{vf(v)}$$by AM-GM, so $vf(v)\le1$. Then $vf(v)+vf(v)\le2$, so by taking $x=v$ we find that $u=v$. This yields a contradiction as $f(v)\le\frac1v$. Suppose $f(u)<\frac1u$, and let $d$ be any positive real. Then $uf(u+du)+(u+du)f(u)>2$ by taking $x=u$ and $y=u+du$, but: $$uf(u+du)+(u+du)f(u)\le uf(u)+(u+du)f(u)<2+d$$since $f$ is decreasing and $f(u)<\frac1u$. Then $uf(u+du)+(u+du)f(u)\le2$, contradiction. The only remaining option is $f(u)=\frac1u$ as desired.

Let $g:\mathbb{R^+} \to \mathbb{R^+}$ be a function such that $g(x)=f(x)/x$ for positive real $x$. Then the given condition is equivalent to : For every $x \in R^+$ there exists exactly one $y \in \mathbb{R^+} $ such that $g(x)+g(y)\leq 2/xy$. Let's call such a pair $(x,y)$ a good pair. Let $G$ be the set of all good pairs. Then obviously $(y,x) \in G$ while $(x,z) \notin G$ for $y\neq z$. 1) $g(x)$ is strictly decreasing. Let $0<a<b$ be positive real numbers. Case 1: $(a,b) \notin G$ Suppose $(b,c) \in G$ . $\therefore$ $(a,c) \notin G$. $\therefore$ $g(a)+g(c)>2/ac$ and $g(b)+g(c) \leq 2/bc$. By combining these two, we have $g(a)-g(b)\geq{\frac{2}{c}}({\frac{1}{a}}-{\frac{1}{b}})>0$. $\therefore$ $g(a)>g(b)$. Case 2: $(a,b) \in G$ Let $d \in \mathbb{R^+}$ such that $a<d<b$. $\therefore$ $(a,d),(d,b)\notin G$ By using the result in Case $1$ we have $g(a)>g(d)$ and $g(d)>g(b)$. $\therefore$ $g(a)>g(b)$ 2) If $(x,y)\in G$ then $g(x)+g(y)=2/xy$ Suppose there exists $(x,y) \in G$ which doesn't hold the above. $\therefore$ $g(x)+g(y)<2/xy$. $\therefore$ $\exists$ $n>0$ so that $g(x)+g(y)<{\frac{2}{x(y+n)}}$. But $g(x)+g(y)>g(x)+g(y+n)$. $\therefore$ $g(x)+g(y+n)<{\frac{2}{x(y+n)}}$ which is a contradiction because $(x,y+n) \notin G$. 3) $g(x)=1/x^2$ and $f(x)=1/x$ for every $x \in \mathbb{R^+}$. We've shown that $g(x)+g(y)\geq2/xy\forall x,y\in\mathbb{R^+}$. $x=y$ $\Rightarrow$ $g(x)\geq1/x^2$. Consider any $(a,b) \in G$. $\therefore$ $2/ab=g(a)+g(b)\geq1/a^2+1/b^2$ $\Rightarrow$ $a=b$. $\therefore$ By step 2, $g(x)=1/x^2.$ $\therefore$ $f(x)=1/x$. It's not difficult to see that $f(x)=1/x$ is indeed a possible solution. ($x/y+y/x\leq2\iff x=y$)

This solution probably is (or close to) the weirdest solution I found on any math Olympiad question. In particular, it hinges on the uncountability of $\mathbb R$. Please help check if it's correct. The only answer is $f(x) = \tfrac 1x$, which satisfies the equation due to AM-GM inequality $\tfrac xy + \tfrac yx \geq 2$. To prove that there are no other solutions, we first substitute $g(x) = xf\left(\tfrac 1x\right)$, turning the inequality to $$g\left(\frac 1x\right) + g\left(\frac 1y\right) \leq \frac{2}{xy} \iff g(x)+g(y)\leq 2xy.$$Our intermediate goal is to show that $f(x)\geq x^2$ for all $x>0$. To do so, consider a very small real number $\delta > 0$ (pick later) and partition $\mathbb R^+$ into strips in form $S_n = [n\delta, (n+1)\delta)$ for $n=0,1,2,\dots$. Consider a strip $S_a$ where $a>1000$. Each number $x\in S_a$ (uncountably infinite numbers) is mapped to another real number $y$ such that $g(x)+g(y)\leq 2xy$. As there are countably many strips, six of them, $a_1,\dots,a_6$ must get mapped $b_1,\dots,b_6$ lying in the same strip $S_b$. Now, note that If $a\ne b$, then just pick the first three $a_1,a_2,a_3$ and $b_1,b_2,b_3$. We have $a_i\ne b_j$ for all $i,j$. If $a=b$, then these six numbers must pair up to at least $3$ pairs. Therefore, we can pick $a_1,a_2,a_3$, $b_1,b_2,b_3$ such that $g(a_i) + g(b_i)\leq 2a_ib_i$ and $a_i\neq b_j$ whenever $i\ne j$. Now, we contend that $a$ and $b$ must be close. Note that in what follows, all constants in the $O$ are absolute, not depending on $\delta$. Claim: $a-b = O(\sqrt a)$. Proof: We have the following table to visualize the argument. \begin{tabular}{c|ccccc} & $g(a_i)$ & & $g(b_i)$ & & $g(a_i)+g(b_i)$ \\[4pt] \hline 1 & $\min$ & + & $\geq(b\delta)^2$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] 2 & $\geq(a\delta)^2$ & + & $\min$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] 3 & $\geq(a\delta)^2$ & + & $\geq(b\delta)^2$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] \end{tabular}Note that since $g(a_i) + g(a_j) \geq 2a_ia_j \geq 2(a\delta)^2$, we have that $g(a_i) \geq (a\delta)^2$ for all but one $i$. Similarly, $g(b_i)\geq (b\delta)^2$ for all but one $i$. Therefore, for some $i$, $g(a_i)\geq (a\delta)^2$ and $g(b_i)\geq (b\delta)^2$. This gives $$\delta^2(a^2+b^2) \leq 2\delta^2(a+1)(b+1) \implies a^2+b^2 \leq 2ab+2a+2b+2,$$giving the desired bound. (Explicitly, $|a-b| < 1+\sqrt{4a+3}$.) $\blacksquare$ Next, we contend that $g(a_1), g(a_2), g(a_3)$ and $g(b_1), g(b_2), g(b_3)$ must be close together. In particular, Claim: We have $g(a_i) = \delta^2(a^2+O(a))$. Similarly, $g(b_i) = \delta^2(a^2+O(a))$. Proof: We first define some notation. Let \begin{align*} m_a &= \min\{g(a_1), g(a_2), g(a_3)\}, \\ M_a &= \max\{g(a_1), g(a_2), g(a_3)\}, \\ m_b &= \min\{g(b_1), g(b_2), g(b_3)\}, \\ M_b &= \max\{g(b_1), g(b_2), g(b_3)\}. \\ \end{align*}Then, we prove that $m_a$ and $m_b$ can't be to small. To do so, we construct the table similar to before \begin{tabular}{c|ccccc} & $g(a_i)$ & & $g(b_i)$ & & $g(a_i)+g(b_i)$ \\[4pt] \hline 1 & $m_a$ & + & $\geq 2(b\delta)^2-m_b$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] 2 & $\geq 2(a\delta)^2-m_a$ & + & $m_b$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] 3 & $\geq 2(a\delta)^2-m_a$ & + & $\geq 2(b\delta)^2-m_b$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] \end{tabular}This time, we have $$2\delta^2(a^2+b^2) - m_a - m_b\leq 2\delta^2(a+1)(b+1) \implies m_a+m_b\geq 2\delta^2(a^2+O(a)).$$Then, we note that $$M_a + m_b \leq g(a_i) + g(b_i) \leq 2\delta^2(a+1)(b+1) = 2\delta^2(a^2+O(a)),$$and analogously, $m_a+M_b \leq 2\delta^2(a^2+O(a))$. Therefore, we have three inequalities \begin{align*} m_a + m_b &\geq 2\delta^2(a^2+O(a)) \\ m_a + M_b &\leq 2\delta^2(a^2+O(a)) \\ m_b + M_a &\leq 2\delta^2(a^2+O(a)). \end{align*}Subtracting the first and second gives $M_b - m_b = \delta^2O(a)$. However, we also have $M_b + m_b \geq 2(b\delta)^2 = 2\delta^2(a^2+O(a))$, so $m_b \geq 2\delta^2(a^2+O(a))$. Analogously, $m_a \geq 2\delta^2(a^2+O(a))$. Therefore, using the second and third inequalities gives the upper bound $M_a,M_b\leq 2\delta^2(a^2+O(a))$. $\blacksquare$ Now, we link this to the entire strip $S_a$. Claim: $g(x) \geq \delta^2(a^2+O(a))$ for all $x\in S_a$. Proof: If $x\in\{a_1,a_2,a_3,b_1,b_2,b_3\}$, we are already done. Otherwise, $$g(x) + g(a_1) \geq 2xa_1 \geq 2\delta^2a^2,$$giving the conclusion as well. $\blacksquare$ By taking $\delta \to 0$ and $a = \left\lfloor\tfrac x\delta\right\rfloor$, we have $$g(x) > \delta^2\left(\frac{x^2}{\delta^2} + O\left(\frac{x}{\delta}\right)\right) = x^2 + xO(\delta),$$yielding $g(x) \geq x^2$ for all $x$. For each $x$, there exists $y$ such that $$2xy \geq g(x)+g(y)\geq x^2+y^2,$$forcing $x=y$ and $g(x)=x^2$, done.

Motivation: You want to have something that lets you perturb $y$ for a given $x$ and use uniqueness to get $f(x)=\dfrac{1}{x}.$ Continuity would be great but this is supposed to be high school level so you naturally try to show monotonicity as the next best thing. Let $x_1>x_2$ and $y\in\mathbb{R}^+$ be the unique number such that $x_1f(y)+yf(x_1)\leq 2.$ Due to the uniqueness of $y$ we have that $x_2f(y)+yf(x_2)>2.$ Since $2\geq x_1f(y)+yf(x_1)>x_2f(y)+yf(x_1)$ and $x_2f(y)+yf(x_2)>2$ we have $f(x_1)<f(x_2).$ Thus the function is decreasing. Now perturbation will give you what you want.

This could have been a 1/4 . Cute little problem though. I claim our answer is $\boxed{f(x) = \tfrac 1x}.$ Define $x \sim y$ if $y$ is the unique value that makes the statement true for $x$. Note that $y \sim x$ too, so this property is commutative. First, we prove $x \sim x$ for all $x$. Suppose FTSoC that $x \sim y$ with $x \neq y$. Then, $x \not \sim x$ and $y \not \sim y$. Hence $2xf(x) > 2$ and similarly $2yf(y) > 2$, so $f(x) > \tfrac 1x$ and $f(y) > \tfrac 1y$. It turns out that\[xf(y) + yf(x) > \frac{x}{y} + \frac{y}{x} > 2\]since AM-GM equality can never hold, contradiction, as desired. So, $x \sim x$ always $\implies f(x) \leq \tfrac{1}{x}$. Now, suppose some $f(c) < \tfrac 1c$, so $f(c) = \tfrac {1}{c + \epsilon}$ for some $\epsilon$. Note that $c \sim c + \epsilon$ then, since\[cf(c + \epsilon) + (c+\epsilon)f(c) = cf(c + \epsilon) + 1 \leq 1 + \frac{c}{c + \epsilon} \leq 2\]which is bad since $c$ can only $\sim$ one number, already established to be itself. Hence, $f(x) = \tfrac 1x$ for all $x$, as desired. $\square$

The only solution is $f(x) = \frac{1}{x}$ for all $x$. AM-GM verifies that $y=x$ is the unique value satisfying $xf(y) + yf(x) \leq 2$ for any $x$. First, we show that $xf(x) \leq 1$ for all $x$. Assume otherwise, then consider $x,y \in \mathbb{R}^+$ with $xf(x) > 1$ satisfying the given inequality. By AM-GM, \[2 \geq xf(y) + yf(x) \geq 2\sqrt{(xf(x))(yf(y))} \implies yf(y) < 1\]Hence, $z=y$ is the unique solution to the inequality \[yf(z) + zf(y) \leq 2\]But we know $z=x$ is also a solution, so $x=y$, and $xf(x) = yf(y) < 1$, contradiction. This implies \[xf(y) + yf(x) > 2 \text{ for all } x \neq y\]Now assume that for some $x$, there is $c < 1$ such that $f(x) = \frac{c}{x}$. Let $y=x+\epsilon$, then \[2 < xf(y) + yf(x) \leq \frac{x}{x+\epsilon} + \frac{c(x+\epsilon)}{x}\]for all $\epsilon\neq 0$. But $\lim_{\epsilon\to 0} RHS = 1 + c < 2$, contradiction. Hence, we are done.

2022 IMO/2 wrote: Let $\mathbb{R}^+$ denote the set of positive real numbers. Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for each $x \in \mathbb{R}^+$, there is exactly one $y \in \mathbb{R}^+$ satisfying \[ xf(y) + yf(x) \le 2 \] The answer is only $\boxed{f(x) = \frac{1}{x}}$. This works, as for $y = x$, we have $2xf(x) = 2$ and for $y \not= x$, by AM-GM, $xf(y) + yf(x) > 2$. We'll now prove that there are no other solutions. Claim 01. For all $x \not= y$, we have \[ \frac{2}{xy} < \frac{f(x)}{x} + \frac{f(y)}{y} \le \frac{1}{x^2} + \frac{1}{y^2} \]Proof. Let us suppose that the unique $y \in \mathbb{R}^+$ satisfying $xf(y) + yf(x) \le 2$ is not $x$. We therefore conclude that $2xf(x) > 2$ and $2yf(y) > 2$, i.e. $f(x) > \frac{1}{x}$ and $f(y) > \frac{1}{y}$. However, this implies \[ \frac{2}{xy} \ge \frac{f(x)}{x} + \frac{f(y)}{y} > \frac{1}{x^2} + \frac{1}{y^2} \ge \frac{2}{xy} \]which is a contradiction. Therefore, we must have $xf(x) \le 1$ for all $x \in \mathbb{R}^+$; and furthermore for all $x \not= y$, we must have \[ \frac{f(x)}{x} + \frac{f(y)}{y} > \frac{2}{xy} \]as claimed. Now, consider the function $g: \mathbb{R}^+ \to \mathbb{R}_{\ge 0}$ such that $g(x) = x^2 - xf \left( \frac{1}{x} \right) \ge 0$ for all $x \in \mathbb{R}^+$. Therefore the condition above rewrites to \[ g(a) + g(b) = a^2 + b^2 - a f \left( \frac{1}{a} \right) - b f \left( \frac{1}{b} \right) < (a - b)^2 \]for any $a \not= b$. Let us write this as $Q(a,b) : g(a) + g(b) < (a - b)^2$. Claim 02. $g \equiv 0$. Proof. We have $0 \le g(a) \le g(a) + g(b) < (a - b)^2$ for all $b \not= a$. Thus, \[ 0 \le g(a) \le \lim_{b \to a} (a - b)^2 = 0 \implies g(a) = 0 \]for any $a \in \mathbb{R}^+$, as desired. This forces $f(x) = \frac{1}{x}$, as desired.

This problem is a joke.

I am struggling more on the IMO 1/4s than the IMO 2/5s.

This problem is quite nice and creative. A good p2!

I think this suits P1 more than P2... I'm disappointed that the uniqueness resolves that quickly for a P2. It's probably nicer as a P1 (compared to 2019 A1?), while P1 could fit as an easy-medium P2, I think. Write $R^+ = \mathbb{R}^+$. For each $x \in R^+$, we claim that the unique $y \in R^+$ satisfying $xf(y) + yf(x) \leq 2$ is $y = x$. In particular, this gives us $x f(x) \leq 1$ and $x f(y) + y f(x) > 2$ for each $x, y \in R^+$. To prove the claim, fix some $x \in R^+$, and suppose that the unique $y$ is distinct from $x$. Then, we have $2 xf(x) \leq 2 \to xf(x) \leq 1$, and similarly $y f(y) \leq 1$. But then $2xy < xy (x f(y) + y f(x)) = x^2 y f(y) + y^2 x f(x) \leq 1 + 1 = 2$, a contradiction. Now, for any $x \in R^+$, denote $g(x) = 1 - x f(x)$. We have $g(x) \geq 0$ for each $x \in R^+$, and also $2xy < xy (x f(y) + y f(x)) = x^2 (1 - g(y)) + y^2 (1 - g(x)) \iff x^2 g(y) + y^2 g(x) < (x - y)^2$ for each $x, y > 0$. For fixed $y$, this means $y^2 g(y) < (y + c)^2 g(y) \leq (y + c)^2 g(y) + y^2 g(y + c) < c^2$ for any $c > 0$. Thus $y^2 g(y) < c$ for any $c > 0$ (for example, by replacing $c$ with $\sqrt{c}$). Since $g(y) \geq 0$ and $y > 0$, this forces $g(y) = 0 \to y f(y) = 1 \to f(y) = \dfrac{1}{y}$ for any $y > 0$. P.S. Writing $\mathbb{R}^+$ as $R^+$ is actually deliberate; I think the above solution should work even if we are working over the set of positive elements of a linear ordered ring $R$. In the case where $R$ is not a field, no such function $f$ exists; otherwise, if $R$ is a field, then $f(x) = 1/x$ is the unique solution. This is why I do not use division at all in the above solution (almost...). An easier solution would start with dividing the original inequality by $xy$ as in #8. One can still prove $y^2 g(y) < c$ for any $c > 0$, for example, as follows: if $c \geq 1$ then $y^2 g(y) < 1^2 \leq c$; otherwise $y^2 g(y) < c^2 < c$.

I wonder if the following variant breaks some solutions. My countability solution still works if we take more $a_i$'s and $b_i$'s. Quote: Determine all $f : \mathbb R^+ \to \mathbb R^+$ such that for each $x\in\mathbb R^+$, there are at least one but at most $2022$ positive real numbers $y$ such that $xf(y)+yf(x)\leq 2$.

Hope this is true Let $y_x>0$ be the unique element for every $x>0$ we have $xf(y)+yf(x)\leq 2$. Assume that $y_x\neq x$ then we have $2xf(x)>2\implies f(x)>\frac{1}{x}$ for some $x>0 (1)$. Then by $AM-GM$: $$2\ge xf(y_x)+y_xf(x)>xf(y_x)+\frac{y_x}{x}\ge 2\sqrt{y_xf(y_x)}\implies f(y_x)<\frac{1}{y_x}$$By $(1)$ we see that $y_{y_{x}}=y_x$, otherwise we get $f(y_x)>\frac{1}{y_x}$, a contracdition. Let $x\to y_x$ then we have $$y_xf(y)+f(y_x)y>2\quad\forall y>0,y\neq y_x$$. But since $x\neq y_x$ so $y_xf(x)+xf(y_x)>2$, a contracditon. This inffer that we get $y_x=x\quad\forall x>0$ and so $$xf(y)+yf(x)>2\quad\forall x,y>0 / x\ne y , \ f(x)\leq \frac{1}{x}$$That is $$\displaystyle\frac{2-\frac{x}{y}}{y}\leq \frac{2-xf(y)}{y}<f(x)\le \frac{1}{x}\quad\forall x,y>0, x\neq y\ (2)$$$$\frac{2-\frac{y}{x}}{x}-\frac{1}{x}<f(y)-f(x)<\frac{1}{y}-\frac{2-\frac{x}{y}}{y}\quad\forall x,y>0,x\neq y$$Thus $\lim_{y\to x} f(y)=f(x)\quad\forall x>0$ so $f$ is continous on $\mathbb{R^+}$. By $(2)$ and squeeze theorem we have $$f(y)=\lim_{x\to y}f(x)=\frac{1}{y}\quad\forall y>0$$So $f(x)=\frac{1}{x}\quad\forall x>0$ and we're done.

I didn't expect this..... Let \(g(x)\) be the unique positive real satisfying the given inequality. I claim that \(f\) is decreasing. Assume that \(f(a)\ge f(b)\) and \(a\ge b\) for some two positive reals, then note that \[g(a)f(b)+bf(g(a))\le g(a)f(a)+af(g(a))\le 2\]contradicting uniqueness of \(g(a)\). Therefore, \(f\) is strictly decreasing. Now, assume that there exists a positive real \(x_0\) for which \(f(x_0)<\frac{1}{x_0}\). Then, clearly \(g(x_0)=x_0\) (since it satisfies the inequality and must be unique). That means \[(x_0+\varepsilon)f(x_0)+x_0f(x_0+\varepsilon)>2\]for \(\varepsilon>0\). Now, since \(f(x_0)<\frac{1}{x_0}\) by assumption, we must have \[\frac{\varepsilon}{x_0}+x_0f(x_0+\varepsilon)>1\]or \[f(x_0+\varepsilon)>\frac{x_0-\varepsilon}{x_0^2}\]And as \(f\) is decreasing: \[f(x_0)>f(x_0+\varepsilon)>\frac{x_0-\varepsilon}{x_0^2}\]so \[x_0^2f(x_0)>x_0-\varepsilon\]which is a contradiction, as we tend \(\varepsilon\rightarrow 0^+\), because \(x_0^2f(x_0)<x_0\). This implies that \(f(x)\ge \frac{1}{x}\) for all \(x\). Assume now that there exists a positive real \(t\) for which \(f(t)>\frac{1}{t}\). Then, \[2\ge g(t)f(t)+tf(g(t))>\frac{g(t)}{t}+\frac{t}{g(t)}\ge 2\]a contradiction. Therefore, we must have \(f(x)=\frac{1}{x}\) for all \(x\). Note that this solution works, since we have the unique positive real satisfying \[\frac{x}{g(x)}+\frac{g(x)}{x}\le 2\]is \(g(x)=x\).

The only answer is $f(x) = \tfrac{1}{x}$. It is easy to prove that if there exists some $t \in \mathbb{R}^+$ with $f(t) < \tfrac{1}{t}$ then there exists $t' \in \mathbb{R}^+$ with $f(t') > \tfrac{1}{t'}$ and vice versa. $\textbf{WLOG}$ suppose there exists $t \in \mathbb{R}^+$ such that $f(t) < \tfrac{1}{t}$. By setting $x = t$ one can find that $y = t$ is the unique $y \in \mathbb{R}^+$ that satisfies $$ tf(y) + yf(t) \leq 2 $$Note that $\tfrac{1}{f(t)} \neq t$, thus $$ tf\left(\frac{1}{f(t)}\right) + \frac{1}{f(t)}f(t) > 2 $$$$ f\left(\frac{1}{f(t)}\right) > \frac{1}{t} > \frac{1}{\frac{1}{f(t)}} = f(t) $$Now we substitute $x = \tfrac{1}{f(t)}$ $$ \frac{f(y)}{f(t)} + yf\left(\frac{1}{f(t)}\right) > tf(y) + yf(t) $$hence the unique $y \in \mathbb{R}^+$ that satisfies $$ \frac{f(y)}{f(t)} + yf\left(\frac{1}{f(t)}\right) \leq 2 $$must be precisely $t$. Therefore $$ \frac{f(t)}{f(t)} + tf\left(\frac{1}{f(t)}\right) \leq 2 $$$$ f\left(\frac{1}{f(t)}\right) \leq \frac{1}{t} $$a contradiction. $\blacksquare$

The only such function is $f(x)=x^{-1}.$ Denote by $X$ the one $y$ of $x.$ If $x>y$ then $yf(X)+Xf(y)>2$ due to uniqueness. Combining implies $f(x)<f(y).$ If $f(x)<x^{-1}$ then $2xf(x)<2$ implies $x=X.$ Then $ f(x+\alpha)<f(x)$ gives $2< xf(x+\alpha)+(x+\alpha)f(x)< xf(x)+\frac{x+\alpha}{x}$ and this is impossible for small $\alpha.$ If $f(x)>x^{-1}$ and $f(X)\geq X^{-1}$ then $2\geq Xf(x)+xf(X) > \tfrac{x}{y}+\tfrac{x}{y}\geq 2$ contradiction. Otherwise $f(X)<X^{-1}$ contradiction again.

Ok but easier than P1 The only solution is $f(x) = 1/x$ which works by AM-GM, we now prove no others work. Let $g(x)$ be the unique value of $y$ such that $xf(y) + yf(x) \le 2$ for a given $x$, and notice that $g(g(x)) = x$. Claim 1. $f$ is strictly decreasing. Proof. Assume FTSOC that there exist $y_1 < y_2$ such that $f(y_1) \le f(y_2)$. We know $g(y_2)f(y_2) + y_2f(g(y_2)) \le 2$. Since $y_1 < y_2$ and $f(y_1) \le f(y_2)$ it follows that $g(y_2)f(y_1) + y_1f(g(y_2)) \le 2$, and therefore $g(y_1) = g(y_2)$. But then $y_1 = g(g(y_1)) = g(g(y_2)) = y_2$, a contradiction. $\blacksquare$. Claim 2. $f(x) \ge 1/x$ for all $x$. Proof. Assume FTSOC that $f(x) = 1/x - c$ for some $c > 0$. Let $y = x + t$ and compute \[xf(y) + yf(x) \le xf(x) + (x + t)f(x) = (2x + t)(1/x - c) = 2 - 2cx + t(1/x - c)\] As $t \to 0$ this goes to $2 - 2cx < 2$, so for all sufficiently small $t$ it is less than $2$, contradicting that $g(x)$ is unique. $\blacksquare$ Now we have $$xf(y) + yf(x) \ge \frac{x}{y} + \frac{y}{x} \ge 2$$ For all $x, y$, with equality if and only if $x = y$ and $f(x) = 1/x$. Since $g(x)$ must exist, this is true for all $x$, and we are done.

Let $g(x)=y$ as per the equation. Claim: $f$ is continuous. Proof: suppose FTSOC that there exists a choice of $x$ and $\epsilon$ such that for all positive real $\delta$, we can find $x^*$ with $|x-x^*|<\delta$ and $|f(x)-f(x^*)|\ge \epsilon$. Then, for a suitably small $\delta$, if $f(x^*)>f(x)$ then $g(g(x^*))$ can take the value of $x$ in addition to $x^*$, and otherwise $g(g(x))$ can take the value of $x^*$ in addition to $x$. (I believe that this part can be adjusted to work even if $g$ has any upper-bounded number of outputs instead of 1) For any fixed $x$, $xf(y)+yf(x)$ is continuous as a function in $y$ since it's the sum of two continuous functions. We can then replace $\le$ by $=$ in the equation. Since we necessarily have $2xf(x)\ge 2$ for all real $x$, we know that $f(x)\ge \frac{1}{x}$. Now, suppose that $g(a)=b$ with $a\neq b$; we have $af(b)+bf(a)\ge \frac{a}{b}+\frac{b}{a}>2$, which is a contradiction, hence $g(x)=x$, and consequentially $f(x)=\frac{1}{x}$.

Unless otherwise specified, all variables in this solution are positive reals. For each $x$, denote the corresponding $y$ by $x^*$. Then, we obviously have $(x^*)^* = x$ and hence $* : x \mapsto x^*$ is a bijection. Note that we have \[ \forall x, \quad xf(x^*) + x^*f(x) \leq 2, \tag*{($1^*$)}\]\[ \forall x, y, \quad y \neq x^* \text{ implies } xf(y)+yf(x) > 2. \tag*{($2^*$)}\] CLAIM 1. $*$ is identity. Proof. Suppose there exists $x$ such that $x^* \neq x$. Now, from $(2)$, we have \[ xf(x) > 1 \quad \text{and} \quad x^*f(x^*) > 1. \]Thus, $(1)$ now gives us \[ 2 \geq xf(x^*) + x^*f(x) \geq 2\sqrt{xx^*f(x)f(x^*)} > 2\]contradiction. $\square$ CLAIM 2. $f(x) = 1/x$ for all $x$. Proof. Now that we know $*$ is identity, $(1)$ and $(2)$ maybe rewritten as follows for the sake of convenience: \[ \forall x \quad f(x) \leq \frac{1}{x}, \tag*{($1^*$)}\]\[ \forall x, y \quad x \neq y \text{ implies } xf(y) + yf(x) > 2. \tag*{($2^*$)}\]Thus, for every $x$ and for every $0 < \varepsilon < x$, we have \[ 2 < xf(x-\varepsilon) + (x-\varepsilon)f(x) \leq \frac{x}{x - \varepsilon} + (x - \varepsilon)f(x). \]This implies that \[ x - 2\varepsilon < (x - \varepsilon)^2 f(x). \]Hence, as $\varepsilon \rightarrow 0$, we have $x \leq x^2f(x)$, that is $f(x) \geq 1/x$. So, we must have $f(x) = 1/x$ for all $x$. It is easy to check if this works since \[ \frac{x}{y} + \frac{y}{x} \leq 2 \Leftrightarrow x = y. \]

The answer is $f \equiv \frac{1}{x}$ for all $x \in \mathbb{R}^{+}$. We say $x \sim y$ if $xf(y)+yf(x) \leq 2.$ We claim that $x \sim y$ iff $x = y.$ If not then $x \not \sim x$ and $y \not \sim y$. So $2xf(x) > 2 \implies f(x) > \frac{1}{x}.$ Similarly, $f(y) > \frac {1}{y}$ as well. Thus, \[xf(y) + yf(x) > \frac{x}{y} + \frac{y}{x} > 2,\] which is a contradiction. Thus we have $x = y$ which implies that $f(x) \leq \frac{1}{x}, \forall x \in \mathbb{R}^{+}$. Now, this means that for $x \neq y,$ we must have $xf(y) + yf(x) > 2$. Take an arbitrary $\epsilon$ and set $y = x + \epsilon.$ Thus, \[xf(x+\epsilon) + (x + \epsilon)f(x) > 2 \implies xf(x+\epsilon) + xf(x) + \epsilon f(x) > 2.\] But we have, \[\frac{x}{x+ \epsilon} + 1 + \epsilon f(x) \geq xf(x+\epsilon) + xf(x) + \epsilon f(x) > 2.\] Hence, \[f(x) > \frac{1}{x + \epsilon}.\] Finally let, $\epsilon \rightarrow 0$ we have $f(x) \geq \frac{1}{x}$ which gives $f(x) = \frac{1}{x}$ as desired.

If $x$ and $y$ satisfy the uniqueness condition then call $x$ and $y$ besties. We wish to show that $x$ must equal $y$ if they are besties. FTSOC assume that $x \neq y$. Then it follows that $x$ and $x$ are not besties, thus $2xf(x) > 2 \implies f(x) > \frac{1}{x}$. However by AM-GM we have $\frac{x}{y} + \frac{y}{x} \geq 2$ so $xf(y) + yf(x) > \frac{x}{y} + \frac{y}{x} \geq 2$, contradiction. Thus $x = y$(and are thus besties). This gives $f(x) \leq \frac{1}{x}$. If $f(x) = \frac{1}{x + \epsilon}$ then we can check that $x$ and $x + \epsilon$ are besties which implies that $\epsilon = 0$. So our only solution is $f(x) = \frac 1x$.

Let $y$ be the buddy of the $x$ is the inequality is true. As a result, both $x$ and $y$ are mutual buddies. we claim the answer is $f(x)=\frac{1}{x}$. Claim: $f(x)\leq \frac{1}{x}$ Proof: FTSOC, assume otherwise. Then: \[LHS>\frac{x}{y}+\frac{y}{x}\geq 2,\]which never works, contradiction $\square$ Claim: $x$ is the buddy of $x$ Proof: FTSOC, assume otherwise, implying that: \[2xf(x)>2,\]\[f(x)>\frac{1}{x},\]contradiction $\square$ Claim: $f(x)\geq \frac{1}{x}$ Proof: FTSOC, assume otherwise. Let $y=\frac{1}{f(x)}$, so the $LHS$ becomes: \[x\cdot f\left(\frac{1}{f(x)}\right)+1<x\cdot f(x)+1<2,\]which implies that $\frac{1}{f(x)}$ is a buddy of $x$. This means that: \[f(x)=\frac{1}{x},\]which is a contradiction $\square$ As a result, $f(x)$ must be $\frac{1}{x}$, so we now prove that it works. To prove that this indeed works, we apply AM-GM, which gives: \[LHS=\frac{x}{y}+\frac{y}{x}\geq 2,\]with equality if $x=y$, implying that the $LHS$ is less than or equal to $2$ if and only if $x=y$, as desired $\blacksquare$

The answer is $f(x) = \tfrac{1}{x}$, which works by AM-GM. Call $y$ a friend of $x$ if $xf(y) + yf(x) \le 2$; then if $y$ is $x$'s only friend, $x$ will be $y$'s only friend. Claim: For all $x$, its only friend is itself. Proof: Suppose ftsoc there exist $x \neq y$ such that $xf(y) + yf(x) \le 2$. Then we have both $xf(x) + xf(x) > 2$ and $yf(y) + yf(y) > 2$. Therefore $f(x) > \tfrac{1}{x}$ and$f(y) > \tfrac{1}{y}$ so $$xf(y) + yf(x) > \frac{x}{y} + \frac{y}{x} > 2$$by AM-GM, contradiction. Thus, since $xf(x) + xf(x) \le 2$, we find that $f(x) \le \tfrac{1}{x}$ for all $x$. Now suppose there exists $x$ such that $f(x) < \tfrac{1}{x}$, so $f(x) = \tfrac{1}{x + \epsilon}$ for some positive $\epsilon$. Then $$xf(x + \epsilon) + (x + \epsilon)f(x) = xf(x + \epsilon) + 1 \le \frac{x}{x + \epsilon} + 1 < 2,$$so $x$ and $x + \epsilon$ are friends, which is a contradiction. Therefore $f(x) = \tfrac{1}{x}$ for all $x$.

Solved with the help of megarnie and abrahms. We claim that the only answer is $f(x) = \dfrac 1x$ and the unique corresponding value of $y$ that satisfies the given condition for any $x$ is $y=x$. We first show that this works. Indeed, we have\begin{align*} xf(y)+yf(x) &= \dfrac{x}{y} + \dfrac yx \\ &\geq 2\sqrt{\dfrac xy \cdot \dfrac yx} \\ &= 2 \end{align*}by AM-GM, which implies that the given inequality is only satisfied when $xf(y)+yf(x) = 2$, and this equality case indeed holds when $x=y$. We now show that $f(x) = \dfrac 1x$ is the only solution. Claim: The given inequality is only satisfied when $y=x$ for any $x \in \mathbb{R^+}$. Since there exists exactly $1$ corresponding value of $y$ for the given inequality to hold for any given $x$, it suffices to show that the given inequality does not hold for any $y \neq x$. For the sake of contradiction, for some fixed value $x_0$, suppose there exists $y=y_0 \neq x_0$ such that the inequality holds. This is the unique value of $y$ that satisfies the inequality, so when $y=x_0$, we must have$$x_0f(x_0) + x_0f(x_0) = 2x_0f(x_0) > 2$$or $x_0f(x_0) > 1$, which is equivalent to $f(x_0) > \dfrac{1}{x_0}$. By symmetry, we also have $f(y_0) > \dfrac{1}{y_0}$. It follows that$$x_0f(y_0) + y_0f(x_0) > \dfrac{x_0}{y_0} + \dfrac{y_0}{x_0} > 2$$where the equality case doesn't occur, but the supposition is that $x_0f(y_0) + y_0f(x_0) \leq 2$, which means $2 < 2$, yielding a contradiction. $\blacksquare$ Then, we have $xf(x) + xf(x) = 2xf(x) \leq 2$, or $f(x) \leq \dfrac 1x \, \forall x \in \mathbb{R^+}$. As we desire to prove that $f(x) = \dfrac 1x \, \forall x \in \mathbb{R^+}$ is the only solution, we alternatively show that there does not exist $x_1 \in \mathbb{R^+}$ such that $f(x_1) < \dfrac{1}{x_1}$. For the sake of contradiction, suppose such $x_1$ exists. Then, we must have $f(x_1) = \dfrac{1}{kx_1}$ for some constant $k>1$. Further, it is established that $f(y) \leq \dfrac 1y \, \forall y \in \mathbb{R^+}$, so for all $y \neq x$, we have $$2 < x_1f(y)+yf(x_1) < \dfrac{x_1}{y} + \dfrac{y}{kx_1}$$Here, to arrive at a contradiction, we may let $y=kx_1 \neq x$, which leads to$$2 < 1 + \dfrac 1k$$and this is impossible since $\dfrac 1k < 1$. $\blacksquare$

If $y$ is the unique number satisfying the condition for $x$ then $x$ must be the unique number satisfying the condition for $y$. Noting this we can see that, if $a$ is paired with $b\neq a$ then $f(a)>1/a$ and $f(b)>1/b$, but then $2\geq af(b)+bf(b)>a/b+b/a$ which is impossible. So $x$ must be always paired with itself, thus $f(x)\leq 1/x$ for all $x$. Now suppose, for some $a$, $a\neq f(a)^{-1}$ then $\frac{a}{f(a)^{-1}}\geq af(f(a)^{-1})+f(a)^{-1}f(a)-1>1$, a contradiction. Thus $f(x)=1/x$ for all $x>0$.

Let the $y$-value of a positive real $x$ be the unique $y$ for which $xf(y) + yf(x) \leq 2$. Claim: the $y$-value of each $x$ is just $x$. Proof: Suppose there exists some $a$ which has a $y$-value of $b \neq a$. So, $af(a) + af(a) > 2 \implies f(a) > \tfrac{1}{a}$. So, \[2 \sqrt{\left( a f(b) \right) \cdot \left( \frac{b}{a} \right)} \leq a f(b) + \frac{b}{a} < a f(b) + bf(a) \leq 2 \implies f(b) < \frac{1}{b},\]which is a contradiction since $b$ has two possible $y$-values, namely itself and $a$. As a consequence of this claim, $f(x) \leq \tfrac{1}{x}$ for all $x$. Claim: $f(x) = \tfrac{1}{x}$. Proof: Suppose there exists some $a$ for which $f(a) < \tfrac{1}{a}$. Then, we have \[ 2 > a + f(a) \geq \tfrac{a}{f(a)} \cdot f(a) + a \cdot f \left( \frac{a}{f(a)} \right),\]so $\tfrac{a}{f(a)} = a \implies f(a) = 1$. Thus, $a < 1$. However, now plugging in $\sqrt{a}$ into the initial assertion gives us a problem: since \[\sqrt{a} f(a) + a f ( \sqrt{a} ) = \sqrt{a} + a f ( \sqrt{a} ) \leq \sqrt{a} + \sqrt{a} < 2,\]$a$ is a $y$-value for $\sqrt{a}$, contradiction.

The only solution is $\boxed{f(x)=1/x}$ which can be seen to work. Claim: The unique value of $y$ is $x$ Assume not that $y\neq x$ satisfied the condition. Then we must have $2<xf(x)+xf(x)$ and $2<yf(y)+yf(y)$ so we can not have $xf(y)+yf(x)\leq 2$. Corollary: $f(x)\leq 1/x$ Claim: $f(x)=1/x$ Assume there exists an $a$ such that $f(a)<1/a$ then choosing a $b$ very close to $a$ yields a contradiction. Formal proof sown in above solutions.

Since there is exactly one $y$ for each $x$ such that $xf(y)+yf(x)\leq 2$, this means that $xf(y)+yf(x)>2$ for every $y$ but one. Now, suppose that for some $x$, the value of $y$ such that $xf(y)+yf(x)\leq 2$ is not equal to $x$, meaning $xf(y)+yf(x)>2$ for $y=x$. This gives $f(x)>\frac{1}{x}$. However, this implies \[xf(y)+yf(x) > \frac{x}{y}+\frac{y}{x}\geq 2\]by AM-GM, meaning there is no $y$ such that $xf(y)+yf(x)\leq 2$, contradiction. Hence, when $y=x$, we must have $xf(y)+yf(x)\leq 2$, giving $f(x)\leq \frac{1}{x}$. Suppose that $f(x)=\frac{1}{x}$ was not true for all $x$, meaning there exists some $a$ such that $f(a)=\frac{1}{a}-c$. Then, for any $x=a+\varepsilon$, with $\varepsilon > 0$ we would have \[2<xf(a)+af(x)=x\left(\frac{1}{a}-c\right)+af(x)\leq \frac{x}{a}-xc+\frac{a}{x},\]which gives us \[\frac{a+\varepsilon}{a}+\frac{a}{a+\varepsilon}-(a+\varepsilon)c>2,\]which can be simplified to \[\frac{\varepsilon^2}{a(a+\varepsilon)^2}>c,\]which we see is clearly false for sufficiently small $\varepsilon$. Thus, we must have $f(x)=\frac{1}{x}$ for all $x$, and it is easy to check this satisfies our inequality conditions by AM-GM.

We will show that $f(x) = \frac{1}{x}$ is the only answer and it is obviously works if x = y. Now let us have f such that each $x \in R^+$ has a friend y with $xf(y) + yf(x) \le 2$. By symmetry it is obvious that y is also the friend of x. Claim: Every number is its own friend. Proof: FTSOC let $a \neq b$ be friends. Then we have that $af(a) + af(a) > 2$ $\Rightarrow$ $f(a) > \frac{1}{a}$. Similarly, $f(b) > \frac {1}{b}$. Using this into the starting inequality we get that $2 \ge af(b) + bf(a) > \frac {a}{b} + \frac {b}{a} \ge 2$ (the last one follows by AM-GM), which is impossible. So now we have that $f(x) \le \frac{1}{x}$ for all x and $xf(y) + yf(x) > 2$ for all $x \neq y$. Let $y = x + \varepsilon$ for $x>0$ and $\varepsilon > 0$. So now we get that $2 < xf(x + \varepsilon) + (x + \varepsilon)f(x) \le \frac{x}{x + \varepsilon} + (x + \varepsilon)f(x)$ $\Rightarrow$ we get that $f(x) > \frac{x + 2\varepsilon}{(x + \varepsilon)^2} = \frac{1}{x + \frac{\varepsilon^2}{x + 2\varepsilon}}$. Since we have that for all $\varepsilon > 0$ this gives us $f(x) \ge \frac{1}{x}$ $\Rightarrow$ $\frac{1}{x} \ge f(x) \ge \frac{1}{x}$ $\Rightarrow$ $f(x) = \frac{1}{x}$ is the only solution, which we already checked that works so we are ready.

MarkBcc168 wrote: This solution probably is (or close to) the weirdest solution I found on any math Olympiad question. In particular, it hinges on the uncountability of $\mathbb R$. Please help check if it's correct. The only answer is $f(x) = \tfrac 1x$, which satisfies the equation due to AM-GM inequality $\tfrac xy + \tfrac yx \geq 2$. To prove that there are no other solutions, we first substitute $g(x) = xf\left(\tfrac 1x\right)$, turning the inequality to $$g\left(\frac 1x\right) + g\left(\frac 1y\right) \leq \frac{2}{xy} \iff g(x)+g(y)\leq 2xy.$$Our intermediate goal is to show that $f(x)\geq x^2$ for all $x>0$. To do so, consider a very small real number $\delta > 0$ (pick later) and partition $\mathbb R^+$ into strips in form $S_n = [n\delta, (n+1)\delta)$ for $n=0,1,2,\dots$. Consider a strip $S_a$ where $a>1000$. Each number $x\in S_a$ (uncountably infinite numbers) is mapped to another real number $y$ such that $g(x)+g(y)\leq 2xy$. As there are countably many strips, six of them, $a_1,\dots,a_6$ must get mapped $b_1,\dots,b_6$ lying in the same strip $S_b$. Now, note that If $a\ne b$, then just pick the first three $a_1,a_2,a_3$ and $b_1,b_2,b_3$. We have $a_i\ne b_j$ for all $i,j$. If $a=b$, then these six numbers must pair up to at least $3$ pairs. Therefore, we can pick $a_1,a_2,a_3$, $b_1,b_2,b_3$ such that $g(a_i) + g(b_i)\leq 2a_ib_i$ and $a_i\neq b_j$ whenever $i\ne j$. Now, we contend that $a$ and $b$ must be close. Note that in what follows, all constants in the $O$ are absolute, not depending on $\delta$. Claim: $a-b = O(\sqrt a)$. Proof: We have the following table to visualize the argument. \begin{tabular}{c|ccccc} & $g(a_i)$ & & $g(b_i)$ & & $g(a_i)+g(b_i)$ \\[4pt] \hline 1 & $\min$ & + & $\geq(b\delta)^2$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] 2 & $\geq(a\delta)^2$ & + & $\min$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] 3 & $\geq(a\delta)^2$ & + & $\geq(b\delta)^2$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] \end{tabular}Note that since $g(a_i) + g(a_j) \geq 2a_ia_j \geq 2(a\delta)^2$, we have that $g(a_i) \geq (a\delta)^2$ for all but one $i$. Similarly, $g(b_i)\geq (b\delta)^2$ for all but one $i$. Therefore, for some $i$, $g(a_i)\geq (a\delta)^2$ and $g(b_i)\geq (b\delta)^2$. This gives $$\delta^2(a^2+b^2) \leq 2\delta^2(a+1)(b+1) \implies a^2+b^2 \leq 2ab+2a+2b+2,$$giving the desired bound. (Explicitly, $|a-b| < 1+\sqrt{4a+3}$.) $\blacksquare$ Next, we contend that $g(a_1), g(a_2), g(a_3)$ and $g(b_1), g(b_2), g(b_3)$ must be close together. In particular, Claim: We have $g(a_i) = \delta^2(a^2+O(a))$. Similarly, $g(b_i) = \delta^2(a^2+O(a))$. Proof: We first define some notation. Let \begin{align*} m_a &= \min\{g(a_1), g(a_2), g(a_3)\}, \\ M_a &= \max\{g(a_1), g(a_2), g(a_3)\}, \\ m_b &= \min\{g(b_1), g(b_2), g(b_3)\}, \\ M_b &= \max\{g(b_1), g(b_2), g(b_3)\}. \\ \end{align*}Then, we prove that $m_a$ and $m_b$ can't be to small. To do so, we construct the table similar to before \begin{tabular}{c|ccccc} & $g(a_i)$ & & $g(b_i)$ & & $g(a_i)+g(b_i)$ \\[4pt] \hline 1 & $m_a$ & + & $\geq 2(b\delta)^2-m_b$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] 2 & $\geq 2(a\delta)^2-m_a$ & + & $m_b$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] 3 & $\geq 2(a\delta)^2-m_a$ & + & $\geq 2(b\delta)^2-m_b$ & = & $\leq 2\delta^2(a+1)(b+1)$ \\[4pt] \end{tabular}This time, we have $$2\delta^2(a^2+b^2) - m_a - m_b\leq 2\delta^2(a+1)(b+1) \implies m_a+m_b\geq 2\delta^2(a^2+O(a)).$$Then, we note that $$M_a + m_b \leq g(a_i) + g(b_i) \leq 2\delta^2(a+1)(b+1) = 2\delta^2(a^2+O(a)),$$and analogously, $m_a+M_b \leq 2\delta^2(a^2+O(a))$. Therefore, we have three inequalities \begin{align*} m_a + m_b &\geq 2\delta^2(a^2+O(a)) \\ m_a + M_b &\leq 2\delta^2(a^2+O(a)) \\ m_b + M_a &\leq 2\delta^2(a^2+O(a)). \end{align*}Subtracting the first and second gives $M_b - m_b = \delta^2O(a)$. However, we also have $M_b + m_b \geq 2(b\delta)^2 = 2\delta^2(a^2+O(a))$, so $m_b \geq 2\delta^2(a^2+O(a))$. Analogously, $m_a \geq 2\delta^2(a^2+O(a))$. Therefore, using the second and third inequalities gives the upper bound $M_a,M_b\leq 2\delta^2(a^2+O(a))$. $\blacksquare$ Now, we link this to the entire strip $S_a$. Claim: $g(x) \geq \delta^2(a^2+O(a))$ for all $x\in S_a$. Proof: If $x\in\{a_1,a_2,a_3,b_1,b_2,b_3\}$, we are already done. Otherwise, $$g(x) + g(a_1) \geq 2xa_1 \geq 2\delta^2a^2,$$giving the conclusion as well. $\blacksquare$ By taking $\delta \to 0$ and $a = \left\lfloor\tfrac x\delta\right\rfloor$, we have $$g(x) > \delta^2\left(\frac{x^2}{\delta^2} + O\left(\frac{x}{\delta}\right)\right) = x^2 + xO(\delta),$$yielding $g(x) \geq x^2$ for all $x$. For each $x$, there exists $y$ such that $$2xy \geq g(x)+g(y)\geq x^2+y^2,$$forcing $x=y$ and $g(x)=x^2$, done. best solution i've seen in a long time

We claim that the only working function is $f(x) = \frac 1x$. We can see this works because we have $$\frac xy + \frac yx \ge 2$$by AM-GM, with equality holding only for $x = y$. Motivated by this solution, we try to find the value $y$ for which $xf(y)+yf(x) \leq 2$ holds (call it the friend of $x$). We claim that the only friend of $x$ is $x$ itself (that is, $x$ is a very lonely person, just like me ). Indeed assume otherwise that $a$ and $b$ are friends, with $a \ne b$. Note that $f(a) > \frac 1a$ and $f(b) > \frac 1b$ by putting $(a,a)$ and $(b, b)$. But putting $(a, b)$, we have $$af(b) + bf(a) > \frac ab + \frac ba \ge 2,$$a contradiction to the fact that $a$ and $b$ are friends. $\square$ Now we effectively have $f(x) \le \frac 1x$ as well as $xf(y) + yf(x) > 2 \forall x \ne y$. The second equation gives $f(y) > \frac 2x - \frac{yf(x)}{x} \ge \frac 2x - \frac y{x^2} \forall x \ne y$ But now taking $\lim_{x \to y}$, we have $f(y) \ge \frac 1y$. Therefore $f(x) \ge \frac 1x \forall x$, and using this with the first equation we have $\boxed{f(x) = \frac 1x}$ as desired. $\square$