Problem

Source: 2016 Taiwan TST Round 1 Quiz 1 Problem 1

Tags: function, algebra, Taiwan, Functional inequality



Suppose function $f:[0,\infty)\to[0,\infty)$ satisfies (1)$\forall x,y \geq 0,$ we have $f(x)f(y)\leq y^2f(\frac{x}{2})+x^2f(\frac{y}{2})$; (2)$\forall 0 \leq x \leq 1, f(x) \leq 2016$. Prove that $f(x)\leq x^2$ for all $x\geq 0$.