This may not be right, but can we not just say: In the inequality, let x=y, giving f(x) >= (x^2)/2. Then since for all x there exists y such that f(x)+f(y)=xy, whatever this y is, it must satisfy xy >= (x^2 + y^2)/2, which by sums of squares being nonnegative implies that y must equal x. Therefore, for all x, f(x)+f(x)=x^2, so f(x)=(x^2)/2 for all x. In substituting this back in we find that it is indeed a solution.

$x=y\implies f(x)\ge x^2/2$ for all $x\in R$. Hence $xy=f(x)+f(y)\ge x^2/2+y^2/2\ge xy$ for all $x\in R$ and certain $y\in R $. Hence we have inequality so $f(x)=x^2/2$ for all $x \in R$. Of course such function satisfies all of the conditions.

Setting $x=y$ in the inequality, we have $f(x)\ge\frac{x^2}2$. Then: $$xy=f(x)+f(y)\ge\frac{x^2+y^2}2\ge\frac{2xy}2=xy$$by AM-GM, so equality holds and $x=y$. From the second equation we have $\boxed{f(x)=\frac{x^2}2}$ which is a solution by AM-GM.

The only solution is $\boxed{f(x) = \frac{x^2}{2}}$. This clearly satisfies the inequality by AM-GM and setting $x = y$ clearly gives $f(x) + f(y) = xy$. Now we prove that it's the only solution. Setting $x = y$ in the inequality gives $f(x) \ge \frac{x^2}{2}$. Now if $f(k) > \frac{k^2}{2}$, then if $f(k) + f(y) = ky$, we have $f(k) + f(y) > \frac{k^2}{2} + \frac{y^2}{2} \ge ky$ by AM-GM, so $f(k) + f(y)$ can never equal $ky$, contradiction. Hence $f(x)$ must equal $\frac{x^2}{2}$ for all $x$.