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BALKAN 2010 Problem 1

  • Paper

The left-hand side is equal to

a3b3+b3c3+c3a3−a3b2c−b3c2a−c3a2b(a+b)(b+c)(c+a)\frac{a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}-a^{3} b^{2} c-b^{3} c^{2} a-c^{3} a^{2} b}{(a+b)(b+c)(c+a)}

so it is enough to show that a3b3+b3c3+c3a3≥a3b2c+b3c2a+c3a2ba^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3} \geq a^{3} b^{2} c+b^{3} c^{2} a+c^{3} a^{2} b. The AM-GM inequality gives us a3b3+a3b3+a3c3≥3a3b3⋅a3b3⋅a3c33=3a3b2ca^{3} b^{3}+a^{3} b^{3}+a^{3} c^{3} \geq 3 \sqrt[3]{a^{3} b^{3} \cdot a^{3} b^{3} \cdot a^{3} c^{3}}=3 a^{3} b^{2} c; summing this inequality and its cyclic analogs yields the desired inequality. Equality holds if and only if a=b=ca=b=c.

Dataset text NuminaMath olympiads. Quoted with source.

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