Functional equation in real numbers - Prove that f(x) = x

Question

The function $f$ is defined over the real numbers and satisfies $f(x) \le x$ and $f(x+y) \le f(x) + f(y)$ for all real $x,y$. Prove that $f(x) = x$ for every real number $x$.

I have tried plugging in some values. Plugging $x=0$ we get $f(0) \le 0$ and $f(0)\le2f(0) \implies f(0)\ge0$ which together imply that $f(0)=0\\$.
To prove the given statement how can I proceed further?

Edit: I tried putting $y=-x$ from which I got
$f(0) = 0 \le f(x) +f(-x)\le x + (-x)=0$
Therefore $f(x)+f(-x)=0 \implies f(-x)=-f(x)$.

How can I use this to show $f(x)=x $ ?

Answer 1

Hint: By hypothesis, $f(-x)\le -x$. But you know that $f(-x)=-f(x)$, so...