<just_to_bump_this_topic> It's quite clear that the linear functions work.. but are there others? </just_to_bump_this_topic>

Linear functions satisfy our condition, and we prove that only these do. If we replace $y$ by $\frac{1}{y}$ we can enhance the condition: \[f(x+\frac{1}{x})+f(y+\frac{1}{y})=f(x+\frac{1}{y})+f(y+\frac{1}{x})=f(x+y)+f(\frac{1}{x}+\frac{1}{y}) \] .Now pick up a fixed $1<y<C$ for some $C>1$ and let $x$ be sufficiently big. Rewriting the second two parts of the condition as \[f(x+y)-f(x+\frac{1}{y})=f(\frac{1}{x}+y)-f(\frac{1}{x}+\frac{1}{y}) \] Now as $f$ is continuous then $f$ is uniformly continuous on $[\frac{1}{C};2C]$ hence for any desired $\epsilon>0$ there exists an $a>0$ such that $|f(x)-f(y)|<\epsilon$ whenever $x,y \in [\frac{1}{C};2C], |x-y|<a$. Thus taking $x>max \{C, \frac{1}{a}\}$ we deduce that $|f(\frac{1}{x}+\frac{1}{y})-f(\frac{1}{y})|, |f(\frac{1}{x}+y)-f(y)|<\epsilon$. This means $(f(\frac{1}{x}+y)-f(\frac{1}{x}+\frac{1}{y}))-(f(y)-f(\frac{1}{}y))<2a$ and we have proven that $\lim_{x\to \infty}(f(x+y)-f(x+\frac{1}{y}))=f(y)-f(\frac{1}{y})$ and the convergence is uniform on any interval $[1;C]$ for $y$. Now any $b>0$ can be written as $y-\frac{1}{y}$ in a unique way for $y>1$. Set $g(b)=f(y)-f(\frac{1}{y})$. Then we see that $g(b)=\lim_{x\to \infty}(f(x+b)-f(x))$. It is from here pretty clear that $g(a+b)=g(a)+g(b)$ and since $g$ is continuous we find that $g(x)=cx$. Now we can suppose that $c=0$ otherwise take $f(x)-cx$ instead of $f$. So we then have $f(x)=f(\frac{1}{x})$ and also $\lim_{x\rightarrow \infty}(f(x+a)-f(x))=0$ uniformly for $a\in [0;C]$. Now take $y$ be fixed and let $x\rightarrow \infty$. The condition $f(x+\frac{1}{x})+f(y+\frac{1}{y})=f(x+\frac{1}{y})+f(y+\frac{1}{x})$ is rewritten as $f(x+\frac{1}{x})-f(x+\frac{1}{y})=f(y+\frac{1}{x})-f(y+\frac{1}{y})$. The RHS now tends to zero according to the limit result obtained just before, and the LHS tends to $f(x+\frac{1}{x})-f(x)$. Thus $f(x+\frac{1}{x})=f(x)$. This is enough to prove that $f$ is constant. Indeed, let $1\leq a<b$. Consider $x_{0}=a, y_{0}=b, x_{i+1}=x_{i}+\frac 1{x_{i}}, y_{i+1}=y_{i}+\frac 1{y_{i}}$. As $x+\frac{1}{x}$ is increasing for $x\geq 1$ and also $(x+\frac{1}{x})^{2}\geq x^{2}+2$ we have $x_{i}<y_{i}$ and $x_{i},y_{i}$ grow infinitely large. Finally we can note that $|x+\frac{1}{x}-y-\frac{1}{y}|<|x-y|$ for $x,y\geq 1$ therefore $|x_{i}-y_{i}|\leq |a-b|$. Therefore $\lim_{n \to \infty}(f(x_{n})-f(y_{n}))=0$. But $f(x_{n})=f(a),f(y_{n})=f(b)$. We conclude that $f(a)=f(b)$ and this finishes the proof.

we can have two real numbers $ a > b > 2$ such that $ f(a) = f(b) = 0$. if f is not a zero function in the interval $ [a,b]$, we can select two real numberes $ a', b'$ such that $ a > a' > b' > b > 2$, $ f(a') = f(b') = 0$, $ f(x) > 0$ for all $ x$ in interval $ (a', b')$. let $ f(c) > 0$ attains the maximum value in $ [a', b']$. It is easy to see that there are two real numbers $ d, e$ in $ [a', b']$ which satisfies $ f(d) + f(c) = f(a) + f(e), (d + c = a + e)$ or possibly $ f(d) + f(c) = f(b) + f(e)$ then, $ f(e) > f(c)$ , which is contradiction. so we have $ f$ a zero fuction in the interval $ [a', b']$. It is easy to extend this interval to whole $ R^ +$

Quote: Find all continuous functions $f(x)$ defined for all $x>0$ such that for every $x$, $y > 0$ \[ f\left(x+{1\over x}\right)+f\left(y+{1\over y}\right)= f\left(x+{1\over y}\right)+f\left(y+{1\over x}\right) . \]Proposed by F. Petrov My solution, mostly copied from here: Replace $x$ by $\frac 1x$ in the given equation, and we have$$f\left(x+\frac1x\right)+f\left(y+\frac1y\right)=f\left(x+y\right)+f\left(\frac1x+\frac1y\right).$$Now take arbitrary positive reals $a,b$ so that $ab\ge 4$, and define the sequences $\left\{a_i\right\}_{i=0}^\infty,\left\{b_i\right\}_{i=0}^\infty$ as follows: $$a_0=a,b_0=b,$$and for $n\ge 0$, take positive reals $x,y$ so that $x+y=a_n,\tfrac1x+\tfrac1y=b_n$ (these exist because $ab\ge 4$), and define $$a_{n+1}=x+\frac1x,b_{n+1}=y+\frac1y.$$ Because of the equation above, $$f(a_0)+f(b_0)=f(a_1)+f(b_1)=\cdots=f(a_n)+f(b_n)=\cdots \quad (\star).$$Also, $a_n+b_n$ is a constant throughout. Now take any $n$, and consider the numbers $a_n,b_n,a_{n+1},b_{n+1}$. Let $x,y$ be the corresponding values that satisfy $x+y=a_n,\tfrac1x+\tfrac1y=b_n.$ This implies $xy=\tfrac{a_n}{b_n}.$ Then we have $$\frac{|a_{n+1}-b_{n+1}|}{|a_n-b_n|}=\left|\frac{x+\frac1x-y-\frac1y}{x+y-\frac1x-\frac1y}\right|=\left|\frac{x-y}{x+y}\right|=\sqrt{1-\frac{4xy}{(x+y)^2}}=\sqrt{1-\frac4{a_nb_n}}.$$But we have $(a_n+b_n)^2\ge 4a_nb_n\implies 1-\frac{4}{a_nb_n}\le 1-\frac{16}{(a_n+b_n)^2},$ so$$\frac{|a_{n+1}-b_{n+1}|}{|a_n-b_n|}\le \sqrt{1-\frac{16}{(a_n+b_n)^2}}=\sqrt{1-\frac{16}{(a_0+b_0)^2}}.$$Now $\textstyle\sqrt{1-\frac{16}{(a_0+b_0)^2}}$ is a constant less than $1$, so $|a_n-b_n|$ converges to zero. It's now easy to see that $a_n,b_n$ are convergent as well, in fact they both converge to $\tfrac{a+b}{2}$; so invoking $(\star )$ and continuity, $$f(a)+f(b)=2f\left(\frac{a+b}{2}\right)\;\forall a,b>0 \text{ so that }ab\ge 4.\quad (\spadesuit)$$Now take arbitrary $a,b>0$. Choose sufficiently large $c,d$ so that: $cd\ge 4$, $bc\ge 4$, $ad\ge 4$, $\frac{a+d}{2}\cdot\frac{b+c}{2}\ge 4$, and $\frac{a+b}{2}\cdot \frac{c+d}{2}\ge 4.$ These conditions are easy to satisfy by taking huge $c,d$; in fact, many of these are redundant. So by applying $(\spadesuit )$ multiple times, we have \begin{align*}f(a)+f(b)+2f\left(\frac{c+d}{2}\right)&=f(a)+f(b)+f(c)+f(d)\\ &=2f\left(\frac{a+d}{2}\right)+2f\left(\frac{b+c}{2}\right)\\ &=4f\left(\frac{a+b+c+d}{2}\right)\\ &=2f\left(\frac{a+b}{2}\right)+2f\left(\frac{c+d}{2}\right).\end{align*}Therefore $f(a)+f(b)=2f\left(\frac{a+b}{2}\right)$ for arbitrary $a,b>0$, and Jensen implies $f(x)=cx+d$, which indeex works. $\blacksquare$