Amir Hossein wrote: Determine all functions $f: \mathbb R \to \mathbb R$ which satisfy the inequality $$f(a^2) - f(b^2) \leq \left( f(a) + b\right)\left( a - f(b)\right),$$for all $a,b \in \mathbb R$. Let $P(x,y)$ be the assertion $f(x^2)-f(y^2)\le (f(x)+y)(x-f(y))$ $P(y,x)$ $\implies$ $f(x^2)-f(y^2)\ge -(f(y)+x)(y-f(x))$ And so $-(f(y)+x)(y-f(x))\le (f(x)+y)(x-f(y))$ Which is new assertion $Q(x,y)$ : $f(x)f(y)\le xy$ $Q(1,-1)$ $\implies$ $f(1)f(-1)\le -1<0$ and so one of $f(1)$ or $f(-1)$ is $>0$ and the other is $<0$ Adding $Q(1,x)$ and $Q(1,-x)$, we get $f(1)(f(x)+f(-x))\le 0$ Adding $Q(-1,x)$ and $Q(-1,-x)$, we get $f(-1)(f(x)+f(-x))\le 0$ And so (see three lines above) $f(-x)=-f(x)$ $Q(x,-y)$ $\implies$ $f(x)f(y)\ge xy$ And so $f(x)f(y)=xy$ and so (setting there $y=1$ and since $f(1)\ne 0$) : $f(x)=ax$ for some real $a$ Plugging this back in originel equation, we get $a^2=1$ and so two solutions : $\boxed{\text{S1 : }f(x)=x\quad\forall x}$ $\boxed{\text{S2 : }f(x)=-x\quad\forall x}$

Amir Hossein wrote: Determine all functions $f: \mathbb R \to \mathbb R$ which satisfy the inequality $$f(a^2) - f(b^2) \leq \left( f(a) + b\right)\left( a - f(b)\right),$$for all $a,b \in \mathbb R$.

This also appeared in Netherlands TST 2018, Exam 3, Problem 2.

Amir Hossein wrote: Determine all functions $f: \mathbb R \to \mathbb R$ which satisfy the inequality $$f(a^2) - f(b^2) \leq \left( f(a) + b\right)\left( a - f(b)\right),$$for all $a,b \in \mathbb R$. Dutch IMOTST

Let $P(a,b)$ be the given assertion. $P(0,0)\Rightarrow f(0)^2\le0\Rightarrow f(0)=0$ $P(x,0)\Rightarrow f(x^2)\le xf(x)$ $P(0,x)\Rightarrow-f(x^2)\le-xf(x)\Rightarrow f(x^2)\ge xf(x)$ So $f(x^2)=xf(x)$, hence $f$ is odd. Now $P(x,y)$ gives the new assertion $Q(x,y):f(x)f(y)\le xy$. $Q(x,-y)\Rightarrow-f(x)f(y)\le-xy\Rightarrow f(x)f(y)\ge xy$ So equality holds in $Q(x,y)$, so $f(x)=\frac1{f(1)}x$ after setting $y=1$. Then $x=1$ yields the solutions $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$, which both work.