2018 USAMO Problems/Problem 1

Problem 1

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

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.

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2018 USAMO Problems/Problem 1

Problem 1

Solution