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Algebra Difficulty 4.6 AIME Prove it Brazil

Prove that
(a+b)(a+c)2abc(a+b+c) (a+b)(a+c) \ge 2\sqrt{abc(a+b+c)}
for all positive real numbers aa, bb and cc.

Solution

By AM-GM,
(a+b)(a+c)=bc+a(a+b+c)2bca(a+b+c) (a+b)(a+c) = bc + a(a+b+c) \geq 2\sqrt{bc \cdot a(a+b+c)}

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