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.

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See also

2018 USAJMO (ProblemsResources)
Preceded by
Problem 1
Followed by
Problem 3
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All USAJMO Problems and Solutions
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