Let $g(x)=x-f(x)$. Then the problem can be restated as follows: Quote: Assume that a function $g\colon\mathbb R\to\mathbb R$ satisfies the following condition: For every $x,y\in\mathbb R$ such that $(g(x)-x-y)(g(y)-y-x)>0$, we have $g(x)=g(y)$. Prove that $g(x)\geq g(y)$ whenever $x>y$. If $g$ is constant then $g$ is nondecreasing, so the problem is proved. From now on we assume $g$ is nonconstant. We also consider the function $h:\mathbb R\to\mathbb R$ by $h(x)=-g(-x)$ for all $x\in\mathbb R$. Then $h$ satisfies the condition on $g$. For a function $\phi\colon\mathbb R\to\mathbb R$ and a number $c\in\mathbb R$, we define $\phi^{-1}\{c\}\doteqdot\{x\in\mathbb R\mid \phi(x)=c\}$. Lemma 1: For every real number $c$, $g^{-1}\{c\}$ is bounded.

Lemma 2: Let $a,b$ be real numbers such that $a<b$ and $g(b)-b<g(a)-a$. Then for all $x\in(g(b)-b,g(a)-a)$ we have either $g(x)=g(a)$ or $g(x)=g(b)$.

Lemma 3: Let $a,b,c$ be real numbers such that $a<b$ and $g(a)=g(b)=c$. Then for all $x\in(c-b,c-a)$ we have $g(x)=c$.

Lemma 4: Let $a,b$ be real numbers such that $a<b$ and $g(a)=g(b)$. Then $g$ is constant on $[a,b]$.

As a conclusion from Lemma 1 and Lemma 4, for every real number $c$, $g^{-1}\{c\}$ is a bounded segment (possibly empty). Lemma 5: Let $c$ be a real number such that $g^{-1}\{c\}$ is nonempty. Let $a=\inf g^{-1}\{c\}$ and $b=\sup g^{-1}\{c\}$. Then $c=a+b$.

Let $x,y$ be real numbers such that $x>y$. Let $I_1=g^{-1}\{g(x)\}$ and $I_2=g^{-1}\{g(y)\}$. Also let $a_1=\inf I_1$, $b_1=\sup I_1$, $a_2=\inf I_2$ and $b_2=\sup I_2$. Since $I_1,I_2$ are segments, $x\in I_1$, $y\in I_2$ and $x>y$, we have $a_1\geq a_2$ and $b_1\geq b_2$. By Lemma 5 we have $$g(x)=a_1+b_1\geq a_2+b_2=g(y)$$as desired.

Let $g(x) = f(x)-x$. Then the condition is \[(g(x)+x+y)(g(y)+x+y)>0\Rightarrow g(x)=g(y)\]and we have to show that $g$ is non-increasing. For every $x>y$, if $y<-g(x)-x$ then $P(x,y)$ gives that $g(x)=g(y)$ or $g(y)+x+y\geq 0$. The latter gives that $g(y)\geq -x-y>g(x)$. Therefore in either way we have $g(y)\geq g(x)$. The same argument holds under the assumption that $x>-g(y)-y$. Therefore everything is done unless \[-g(x)-x\leq y < x\leq -g(y)-y.\]If this occurs, then we will have $g(y)<-2y$ and $g(x)>-2x$. Now suppose that $t>s$ and $g(t)>g(s)$. By Step 2. we know that \[-g(t)-t\leq s<t\leq -g(s)-s.\]For any $-g(t)-t<u<-g(s)-s$, from $P(t,u),P(s,u)$ we have that one of the three cases occur: (1) $g(u)=g(t), g(u)\geq -u-s$; (2) $g(u)=g(s), g(u)\leq -u-t$; (3) $-u-s \leq g(u) \leq -u-t$. However, the third case contradicts the assumption that $t>s$. Therefore for every $-g(t)-t<u<-g(s)-s$ we have $g(u)=g(t)$ or $g(s)$. Moreover, if $u<-g(t)-s$ then $g(u)=g(s)$, and if $u>-g(s)-t$ then $g(u)=g(t)$. Now suppose for the sake of contradiction that there are $x>y$ such that $g(x)>g(y)$. By step 3. we have that for any $-g(x)-x<u<-g(x)-y$ we have $g(u) = g(y)$. Take $t=x$ and $s=u$, and then we get that for any $-g(x)-x<u'<-g(x)-u$ we have $g(u')=g(u)=g(y)$. Since we can take $u$ arbitrarily closed to $-g(x)-x$, we have that for any $-g(x)-x<u'<-g(x)-(-g(x)-x)=x$ we have $g(u')=g(y)$. Since $-g(x)-x\leq y$, we have that for any $y<u'<x$ we have $g(u')=g(y)$. Similarly we can get that for any $y<u'<x$ we have $g(u')=g(x)$, which is apparently a contradiction. Therefore the assumption that there exist $x>y$ such that $g(x)>g(y)$ must be false.

A proof using only a picture on a napkin. Put $-f(x)$ instead of $f(x)$. In other words, we denote $g(x)=-f(x)$ , then translate the statement into the terms of $g(\cdot)$ and rename $g$ again as $f$. (all that because I had already drawn that beautiful drawings, using only paint , when realized it should be $g$ instead of $f$ ) Now the problem statement looks like: Quote: A function $f:\mathbb{R} \to \mathbb{R}$ has the following property: $$\text{For every } x,y \in \mathbb{R} \text{ such that }(f(x)-y)(f(y)-x) > 0, \text{ we have } f(x)-y = f(y)-x.$$Prove that $f(x)+x$ is monotone increasing, i.e. $f(y)+y \leq f(x)+x$ whenever $y<x$. For the sake of contradiction, suppose there exist $x,y;\,y<x$ such that $f(y)+y>f(x)+x$, which is equivalent to $f(y)-x>f(x)-y$. Due to the given condition, it's not possible $f(y)-x$ and $f(x)-y$ to be with the same signs. The only possibility is $f(x)\le y<x\le f(y)$. From now on look at the attached drawing. The upper-left picture shows an impossible situation. it cannot happen. The two upper-right drawings show that if $x,y,f(x),f(y)$ are disposed as shown then the distance between $x,y$ equals the distance between $f(x),f(y)$. That's enough to proceed further. Look at drawing $2$. We take some $u$ slightly bigger than $f(x)$ (it's enough to take $u$ such that $u-f(x)<x-y$). Using the above properties of $f$, the only possible position of $f(u)$ is on the right side of $f(y)$ and satisfying $f(u)-f(y)=y-u$. Now, we take some $z$ between $f(y)$ and $f(u)$ and try to figure out where should be its image $f(z)$. Using the two observations, shown on drawing $1$, there is no place where to pin $f(z)$. Game over!

Let $g(x) = x-f(x)$. The original problem becomes $$(x+y-g(x))(x+y-g(y)) > 0 \implies g(x) = g(y) \qquad \qquad \hdots \hdots (0)$$ Claim 1: If $g(a_1) = g(a_2) =:c$ and $a_2 < a_1$ then $y \in (c-a_1, c-a_2) \implies g(y) = c$. Proof: Suppose otherwise. Then from the problem statement, $y> c-a_1$ and letting $x = a_1$ in $(0)$, we get $$(a_1 + y - c)(a_1 + y - g(y)) \le 0 \implies a_1 + y - g(y) \le 0 \implies y-g(y) \le -a_1.$$Likewise $y-g(y) \ge -a_2$. So $$-a_2 \le y-g(y) \le -a_1$$but we've assumed $a_2 < a_2$, contradiction. If everything from the above is true...... Claim 2: [graph of $g$ resembling step function] If $g(a_1) = g(a_2)$ then $g(x) = g(a_1) = g(a_2)$ for all $x \in (a_2,a_1)$. Proof: WLOG $|c/2 - a_1| \ge |c/2 - a_2|$ (the other case is similar). Then Claim 1 tells us that for all $y = c-a_1 - \epsilon$ with $\epsilon > 0$ and sufficiently small, $g(y) = g(a_1)$. Take such $y$ and apply Claim 1 with $a_2$ replaced by $y$ proves Claim 2. In view of Claim 2 let us divide the domain $\mathbb R$ into a family $\mathcal I$ of disjoint intervals such that for each value $\alpha$ in the image of $g$ there exists $I_\alpha \in \mathcal I$ whose image is precisely $\alpha$. We say a number is boxed if it is contained in an interval consisting of more than one number (i.e. a box). We can sharpen the result of Claim 2: since Claim 1 essentially tells us that reflection of a part of a box (with endpoints removed) is still within the box, it follows that any box has its $g$-value as the sum of its endpoints. Claim 3: The problem is true. Suppose otherwise, i.e. $\exists \ y,x$ with $g(y) < g(x)$ and $y > x$. From the sharpening of Claim 2 if both $x,y$ are boxed then we immediately deduce from $y> x$ that $g(y) > g(x)$. By our assumptions $y-g(y) > x-g(x)$ so there exists $z \in (g(y) - y, g(x) - x)$ and if for one such $z$ we have $g(z) \ne g(x), g(y)$ then $$\begin{cases} (z+y-g(z))(z+y-g(y)) \le 0\\(z+y-g(z))(z+y-g(y)) \le 0\end{cases} \implies y \le x,$$false. Thus $g(z) = g(x)$ or $g(y)$ for some $z\ne x,y$ which means that one of $x,y$ (but not both) is boxed, hence exactly one of $x,y$ (say $x$) has its box containing $(g(y)-y, g(x) -x)$. Now $g(x) - y < g(x) - x$ and $g(x) - y > g(y) - y$ so $g(g(x) - y + \epsilon) = g(x)$ for all $\epsilon$ sufficiently close to $0$. Replacing $x$ by $g(x) - y + \epsilon$ for this particular $x$ in $(0)$ gives $$\epsilon(x+\epsilon+y-g(y)) \le 0$$for any $\epsilon \in (-\delta,\delta)$ for some $\delta$ sufficiently close to $0$. The above is a quadratic that changes sign around $\epsilon = 0$ so it must be identically $0$, which is impossible since it has leading coefficient $1$, a contradiction, so we are done.

The problem condition rewrites as: $x-f(x)\ge y-f(y)$ when $x>-f(y)$, and $x-f(x)\le y-f(y)$ when $x<-f(y)$. Now, suppose we have $a,b$ such that $a>b$ but $a-f(a)<b-f(b)$. Then, we must have $b\ge -f(a)$, $a\le -f(b)$, so $-f(a)\le b<a\le -f(b)$. So, if there exists any $c$ with $b<c<a$, we must have $a-f(a)\le c-f(c)\le b-f(b)$. So, if $c-f(c)\not\in\{a-f(a),b-f(b)\}$, then $a$ forms an inversion with $c$, and $f(c)\ge a$. However, $b$ forms an inversion with $c$, so $f(c)\le b$. This is clearly a contradiction, so $c-f(c)\in\{a-f(a),b-f(b)\}$ for all $c\in(a,b)$. WLOG that $\frac{a+b}{2}-f\left(\frac{a+b}{2}\right)=b-f(b)$. Then, we must have $b<\frac{a+b}{2}<a\le -f\left(\frac{a+b}{2}\right)<-f(b)$ Every number $d\in\left(-f\left(\frac{a+b}{2}\right),-f(b)\right)$ satisfies $b-f(b)\ge d-f(d)\ge \frac{a+b}{2}-f\left(\frac{a+b}{2}\right)$, so $d-f(d)=b-f(b)$ for all $d$ in this range. Choose $d_0=-\frac{f\left(\frac{a+b}{2}\right)+f(b)}{2}$. We have $b<-f(d_0)<\frac{a+b}{2}<a$ since $d_0-f(d_0)=b-f(b)$. So, $a$ is strictly in between $-f(d_0)$ and $-f(b)$. So, $d_0-f(d_0)\le a-f(a)\le b-f(b)\implies a-f(a)=b-f(b)$, which is a contradiction, as desired.

Remark: My first A8!

Let $g(x)=x-f(x)$, the condition becomes $(x+y-g(x))(x+y-g(y)) > 0 \implies g(x)=g(y)$ Actually, we can rephrase the condition as follows, by contrapositive: $g(x)< g(y) \implies (x+y-g(x))(x+y-g(y)) \leq 0 \iff x+y \in \big[g(x),g(y)\big]$ And we want to show $g$ is nondecreasing. Assume there exist reals $u,v,a,b$ with $u>v$ and $a=g(u)<g(v)=b$. The condition then gives $b\geq u+v \geq a$, from which we obtain $b-v\geq u > v \geq a-u$. Now let's look at what values $f(t)$ can take for some $t$. $g(t)>b \implies v+t \in \big[b,g(t)\big] \implies t \geq b-v$ $g(t)=b \implies g(t)>a \implies t+u \in \big[a,b \big] \implies t \in \big[ a-u,b-u \big]$ $b>g(t)>a \implies t+u \in \big[a,g(t) \big]$ and $ t+v \in \big[g(t),b \big] \implies t+u \leq g(t) \leq t+v \implies u\leq v$, contradiction $g(t)=a \implies g(t) < b \implies t+v \in \big[a,b \big] \implies t \in \big[ a-u,b-u \big]$ $a>g(t) \implies u+t \in \big[g(t),a \big] \implies t \leq a-u$ A very simple case analysis then gives the following: $t< a-u \implies g(t)< a$ $t\in \big( a-u, a-v \big) \implies g(t)=b$ $t \in \big[a-v,b-u \big] \implies g(t) \in \{a,b \}$ $t\in \big( b-u, b-v \big) \implies g(t)=a$ $t > b-v \implies g(t) > b$ Now, one of the two inequalities $a-u \leq v$ and $b-v \geq u$ must be strict, or else $a=u+v=b$, and let's suppose it's $a-u < v $ (the other case can be handled in the same way). This means we can select a $v' \in \big( a-u, min\{v, a-v\} \big)$, and we know that $g(v')=b$, so we can repeat everything we've said so far with $v'$ replacing $v$, and we obtain $t\in \big( b-u, b-v' \big) \implies g(t)=a$. But this is a contradiction, since for any $k \in \big(b-v,b-v' \big)$ we should have $g(k)>b$ and $g(k)=a$ at the same time. $\square$

Let $g(x) = f(x) - x \text{ } ( \forall x \in \mathbb{R})$. Problem's equation becomes: If: $$ (g(x) + (x + y))(g(y) + (x + y)) > 0 \text{ } (*)$$then: $$g(x) = g(y)$$For a condtradiction let's say $\exists a, b \in \mathbb(R): \begin{cases} a > b \\ g(a) > g(b) \end{cases} $ (here $g(a) > g(b)$ is the same as $f(a) + b > f(b) + a$) Then notice that $ - g(a) - a < -g(b) - b$ . Let's say $\exists x \in [-g(a) - a; -g(b) - b]: g(x) \neq \begin{cases} g(a) \\ g(b) \end{cases}$ Then the $(*)$ doesn't hold for $(a, x)$ and $(b, x)$ so: $$ 0 \leq g(x) + x + b < g(x) + x + a \leq 0 $$Which is a contradiction, so $\forall x \in [-g(a) - a; -g(b) - b]: g(x) = \begin{cases} g(a) \\ g(b) \end{cases}$ Also notice that if $ g(x) = g(b) \neq g(a) $ then: $$ \begin{cases} (g(x) + (x + a))(g(a) + (x + a))\leq 0 \\ g(x) + (x + a) = g(b) + (x + a) < g(a) + (x + a) \end{cases}$$so: $$ \begin{cases} g(x) + (a + x) \leq 0 \\ g(a) + (a + x) \geq 0 \end{cases} $$which is the same as $ -g(b) - a \geq x \geq -g(a) - a$. Doing the same for case $g(x) = g(a)$, we get: $$ \begin{cases} \text{if } x \in [ - g(a) - a; -g(b) - b ]: g(x) = \begin{cases} g(a) \\ g(b) \end{cases} \\ \text{if } g(x) = g(b): x \in [ - g(a) - a; -g(b) - a] \\ \text{if } g(x) = g(a): x \in [ - g(a) - b; -g(b) - b] \end{cases} $$So $ \begin{cases} (-g(a) - a) \in [-g(a) - a; -g(b) - b] \\ (-g(a) - a) \notin [-g(a) - b; - g(b) -b] \end{cases} \Rightarrow g (-g(a) - a) = g(b)$ and $ \forall x < -g(a) - a: x \notin [-g(a) - a; -g(b) - a] \Rightarrow g(x) \neq g(b)$. In layman's terms: $ (-g(a) - a) $ is the smallest number $(x)$ with $g(x) = g(b)$. Same is true for $(-g(b) - b)$ (that it is the biggest number $(x)$ with $g(x) = g(a)$). Notice that $\begin{cases} -g(b) - b > -g(a) - a \\ g(-g(b) - b) = g(a) > g(b) = g(-g(a) - a) \end{cases} $ so if we apply the same logic to $(-g(b) -b; -g(a) - a$ as we applied for $(a, b)$ we get that: $-g(-g(b) - b) - (-g(b) - b)$ is the smallest number $(x)$ for which $g(x) = g(b)$ so: $$ -g(-g(b) - b) - (-g(b) - b) = -g(a) + g(b) + b = -g(a) - a$$$$ g(b) = - (a + b) $$On the other hand, by appling the same logic for $g(a)$: $$ g(a) = - (a + b) $$So we get $$ g(b) < g(a) = -(a + b) = g(b) \text{ which is a condradiction.}$$

This is the first A8 that I solve and spend a lot of time (6+ hours), so I decided to post it on the thread. Let $g(x) = x - f(x)$. Then the problem becomes $(x + y - g(x))(x + y - g(y)) > 0$ then $g(x) = g(y)$ (1). It's enough to prove that g is nondecreasing. We will prove that for all reals $x , y$ if $x \leq y$ , then $g(x) \leq g(y)$. Fix $x$ and all real $y > x$ we will prove $g(x) \leq g(y)$. An real number $x$ is called good if $x > g(x) - x$ , bad if $x < g(x) - x$ , special if $x = g(x) - x$. Claim 1 : For all real $a$ if $b < g(a) - a$ then $g(b) \leq g(a)$ , if $b > g(a) - a$ then $g(b) \geq g(a)$. Proof : If $b < g(a) - a$ and $a < g(b) - b$ , then (1) yields $g(a) = g(b)$ . Then we can assume $a > g(b) - b$. Therefore $g(a) > a + b \geq g(b)$ tells us $g(a) > g(b)$. If $y > g(a) - a$ and $a > g(b) - b$ , then (1) yields $g(a) = g(b)$. Similarly we can assume $a < g(b) - b$. Therefore $g(a) < a + b \leq g(b)$ tells us $g(b) > g(a)$. Thus claim 1 proved. If $x$ if not bad then claim 1 tells us for all real $y > x$ , we have $g(y) \leq g(x)$. Then we can assume that $x$ is bad. Claim 2 : If $x$ is bad, then on the interval $\big[x , g(x) - x)$ $g$ attains at most two values. And similarly $x$ is good , then on the interval $(g(x) - x , x\big]$ $g$ attains at most two values. Proof : Take any $y \in \big[x , g(x) - x)$. Since $y < g(x) - x$ , if $x < g(y) - y$ then (1) yields $g(x) = g(y)$. Thus we can assume $x \geq g(y) - y$ , then $g(x) > x + y \geq g(y)$. Suppose there are $z \in \big[x , g(x) - x)$ such that $g(z) \neq g(x)$ and $g(z) \neq g(y)$. WLOG we can assume $z > y$. Since $g(z) \neq g(x)$ then we have $x \geq g(z) - z$. It's obvious that $y , z$ are good numbers , because $y > x \geq g(y) - y$ and $z > x \geq g(z) - z$. Therefore $z > y > g(y) - y$ and since $y > x \geq g(z) - z$ (1) yields $g(y) = g(z)$ , a contradiction. Similarly in case $x$ is good , then on the interval $(g(x) - x , x\big]$ $g$ attains at most two values. Claim 2 proved. If for all $y \in \big[x , g(x) - x) $ , we have $g(x) = g(y)$ then we are done. Thus we assume there are $y \in \big[x , g(x) - x)$ such that $g(y) \neq g(x)$. Note that $g(y) < g(x)$ and $y$ is good. We assume $g(y) - y < x$ ($g(y) - y = x$ case will be explained below). Since $y$ is good , by claim 2 yields on the interval $(g(y) - y , y \big]$ $g$ attains at most two values , namely $g(x) , g(y)$. Take $a \in (x , y)$. If $g(a) = g(x)$ , then $g(a) - a < g(x) - x$. Notice that $a > g(y) - y$ and if $y > g(a) - a$ then (1) yields $g(y) = g(a) = g(x)$ ,we are done. Thus $y \leq g(a) - a$. Since all real $z > g(a) - a$, we have $g(z) \geq g(a) = g(x)$. Therefore $g$ is constant on interval $(g(a) - a , g(x) - x)$ and arbitrary $k \in (g(a) - a , g(x) - x)$ ,we have $g(k) = g(x)$. Take $k \in (g(a) - a , g(x) - x)$ such that k is close enough to $g(x) - x$. Since $k > g(a) - a \geq y > g(y) - y$ if $y > g(k) - k$ then (1) tells us that $g(y) = g(k) = g(x)$ , we are done. Therefore $y \leq g(k) - k$, or equivalently $y + k \leq g(x)$. Since $k$ is near enough to $g(x) - x$ and $y > x$ , we have $y + k > g(x)$, a contradiction. Therefore for all $a \in (x , y)$ , we have $g(a) = g(y)$. Notice that $x \geq g(a) - a$, otherwise (1) yields $g(x) = g(a) = g(y)$ ,so we are done. Since all real $z < g(a) - a$ , we have $g(z) \leq g(a) = g(y)$, and for all $z \in (g(y) - y , y \big] , g(z) \geq g(y)$. Therefore $g$ is constant on interval $(g(y) - y , g(a) - a)$ and arbitrary $k \in (g(y) - y , g(a) - a)$ , we have $g(k) = g(y)$. Take $k \in (g(y) - y , g(a) - a)$. Since $k < g(a) - a < g(y) - x < g(x) - x$ if $x < g(k) - k$ then (1) yields $g(y) = g(k) = g(x)$, so we are done. Thus $x \geq g(k) - k$ ,or equivalently $x + k \geq g(y)$. But $g(y) \leq x + k < g(y) - y + x < g(y)$, a contradiction. Now only remaining case is $x = g(y) - y$. Take $a \in (x , y)$. If $g(a) = g(x)$, then we can done as above. Therefore for all $a \in (x , y \big] , g(a) = g(y)$. Take arbitrary $a \in (x , y)$. Since $a < g(x) - x$ then if $x < g(a) - a$ (1) yields a contradiction. Thus $g(y) - y = x \geq g(a) - a$ , or equivalently $a > y$ a contradiction. So $g$ is constant on interval $\big[x , g(x) - x)$ , and take arbitrary $a \in \ (x , g(x) - x)$ we have $g(a) = g(x)$ and $g(a) - a < g(x) - x$, therefore $g(x) = g(a) < g(g(x) - x)$. Therefore for all $y \geq x$, we have $g(y) \geq g(x)$ , as desired.