Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$
f\left(a^{2}\right)-f\left(b^{2}\right) \leqslant(f(a)+b)(a-f(b)), \quad \text { for all } a, b \in \mathbb{R}
$$
By taking $(a, b)=(0,0)$, we obtain $f(0)^{2} \leqslant 0$ hence $f(0)=0$. By taking $(a, b)=(x, 0)$ and then $(a, b)=(0, x)$, we find that $f\left(x^{2}\right)=x f(x)$. By substituting into the initial equation, we deduce that $a f(a)-b f(b) \leqslant(f(a)+b)(a-f(b)$, thus $f(a) f(b) \leqslant a b$ for all $a, b$. On the other hand, $f\left(x^{2}\right)=x f(x)$ implies that $x f(x)=-x f(-x)$, so $f$ is odd. By replacing $b$ with $-b$ in $f(a) f(b) \leqslant a b$, we deduce that $f(a) f(b)=a b$ for all $a$ and $b$. In particular, $f(1)^{2}=1$ so $f(1)=1$ or $f(1)=-1$. Since $f(a) f(1)=a$ for all $a$, we conclude that $f(x)=x$ or $f(x)=-x$ for all $x$.
Conversely, it is easily verified that the functions $f(x)=x$ and $f(x)=-x$ are solutions to the functional equation.