# 2018 USAJMO Problems/Problem 2

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## Problem

Let $a,b,c$ be positive real numbers such that $a+b+c=4\sqrt[3]{abc}$. Prove that $$2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.$$

## Solution 1

WLOG let $a \leq b \leq c.$ Add $2(ab+bc+ca)$ to both sides of the inequality and factor to get: $$4(a(a+b+c)+bc) \geq (a+b+c)^2$$ $$\frac{4a\sqrt[3]{abc}+bc}{2} \geq 2\sqrt[3]{a^2b^2c^2}$$

The last inequality is true by AM-GM. Since all these steps are reversible, the proof is complete.