Home Competitions BALKAN 2010 Problem 1
The left-hand side is equal to
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 ) \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)} ( a + b ) ( b + c ) ( c + a ) 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
so it is enough to show that 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^{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 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 . The AM-GM inequality gives us a 3 b 3 + a 3 b 3 + a 3 c 3 ≥ 3 a 3 b 3 ⋅ a 3 b 3 ⋅ a 3 c 3 3 = 3 a 3 b 2 c a^{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 a 3 b 3 + a 3 b 3 + a 3 c 3 ≥ 3 3 a 3 b 3 ⋅ a 3 b 3 ⋅ 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 = c a=b=c a = b = c .
Dataset text NuminaMath olympiads . Quoted with source.
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