For any $x$ we have $f(x) \geq 2xy - f(y)$ for all $y$, hence, for $y=x$, $f(x) \geq x^2$. Define $g(x) = f(x) - x^2 \geq 0$. The relation becomes $0 \leq g(x) = \max\{-g(y) - (x-y)^2 \mid y\in \mathbb{R}\} \leq 0$, therefore $g(x) = 0$, and so $f(x) = x^2$, for all $x$.

actually it is very old problem however here is my solution!) $ f:\mathbb{R}\rightarrow\mathbb{R} $ $ \forall x\in\mathbb{R}\ \ f(x) = max(2xy-f(y)) $ it easy to note that $f(x)\ge 2xy-f(y) \forall x,y\in\mathbb{R}$ by plugging $y=x=>2x^2-f(x)\le f(x)=>f(x)\ge x^2$ consider that there exists such $x_{0}$ that $f(x_{0})>x_{0}^2=>f(x_{0})=x_{0}^2+a$ where $a>0$ $f(x_{0})=2x_{0}y_{0}-f(y_{0})=>f(y_{0})=2x_{0}y_{0}-x_{0}^2-a<2x_{0}y_{0}-x_{0}^2=>f(y_{0})+(x_{0}-y_{0})^2<y_{0}^2$ which is condraction!) thus the answer is $f(x)=x^2$

Why does $f(y_{0})+(x_{0}-y_{0})^2<y_{0}^2$ gives you a contradiction arshakus? arshakus wrote: consider that there exists such $x_{0}$ that $f(x_{0})>x_{0}^2$ I think you should suppose that for all $x$ we have $f(x)> x^2$ in order to get a contradiction.

hatchguy wrote: Why does $f(y_{0})+(x_{0}-y_{0})^2<y_{0}^2$ gives you a contradiction arshakus? because $y_{0}$ is such a number for what $f(y_{0})>y_{0}^2$ hatchguy wrote: I think you should suppose that for all $x$ we have $f(x)> x^2$ in order to get a contradiction. if you suppose for all $x$ we have $f(x)> x^2$ then you must look in case if for some $x$ $f(x)>x^2$ and for others $f(x)\le x^2$.

From $ f(y_{0})+(x_{0}-y_{0})^{2}<y_{0}^{2} $ you have that $f(y_{0}) < y_{0}^2$. Where do you have that $y_{0}^2 < f(y_{0})$?

arshakus wrote: by plugging $y=x=>2x^2-f(x)\le f(x)=>f(x)\ge x^2$ here you are)))