MEMO 2021 T-1

Let $P(a,b)$ denote the assertion for this functional inequality. $P(0,0)\implies (f(0))^2\leq 0$ and hence, $f(0)=0$. Now, $P(a,a)\implies (f(a))^2\leq a^2$. $P(0,a)$ and $P(a,0)$ gives $a\cdot f(a)=f(a^2)$ which basically gives $f(a)=-f(-a)$. $\newline$ Now, Adding $P(a,-a)$ and $P(-a,a)$ gives $-2a^2\geq 2\cdot f(a)\cdot f(-a)=-2(f(a))^2\geq -2a^2\implies (f(a))^2=a^2\implies f(a)=\pm a$. Hence, $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$ are the solutions.

somebodyyouusedtoknow wrote: Determine all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying \[ f(a^2) - f(b^2) \leq (f(a)+b)(a-f(b)) \]for all $a,b \in \mathbb{R}$. https://artofproblemsolving.com/community/c6h1440672p8193521 https://artofproblemsolving.com/community/c6h1530290p9199305 https://artofproblemsolving.com/community/c6h1663424p10565263

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.