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USAMO 2018 (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)\geq a^2+b^2+c^2

.

Core

Toolkit 28 — Max-QM-AM-GM-HM-Min Inequality

Major

Toolkit 24 — Square of a sum (n terms)

WLOG

a\leq b\leq c

2(ab+bc+ca)+4\min(a^2,b^2,c^2)=2(ab+bc+ca)+4a^2

We need to prove

2(ab+bc+ca)+4a^2\geq a^2+b^2+c^2

Add

2(ab+bc+ca)

to both sides.

\Longleftrightarrow

4(ab+bc+ca)+4a^2\geq a^2+b^2+c^2+2(ab+bc+ca)=(a+b+c)^2

By the assumption of the problem,

a+b+c=4\sqrt[3]{abc}

4(ab+bc+ca)+4a^2=4a(a+b+c)+4bc\geq (a+b+c)^2=(4\sqrt[3]{abc})^2

\Longleftrightarrow

4a(4\sqrt[3]{abc})+4bc\geq16\sqrt[3]{a^2b^2c^2}

\Longleftrightarrow

16a\sqrt[3]{abc}+4bc\geq16\sqrt[3]{a^2b^2c^2}

\Longleftrightarrow

4a\sqrt[3]{abc}+bc\geq4\sqrt[3]{a^2b^2c^2}

By Toolkit 28 — Max-QM-AM-GM-HM-Min Inequality

\frac{4a\sqrt[3]{abc}+bc}{2}\geq\sqrt{4a\sqrt[3]{abc}\cdot bc}

\Longrightarrow

4a\sqrt[3]{abc}+bc\geq4\sqrt{ab c\sqrt[3]{abc}}

\Longrightarrow

4a\sqrt[3]{abc}+bc\geq4\sqrt[3]{a^4b^4c^4}

\Longrightarrow

4a\sqrt[3]{abc}+bc\geq4\sqrt[3]{a^2b^2c^2}

AMC 12A 2023 (Problem 23)
AMC 10B Fall 2021 (Problem 22)
USAJMO 2023 (Problem 1)
AMC 12A 2024 (Problem 15)

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