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Brazilian Mathematical Olympiad

Algebra Difficulty 4.6 AIME Prove it Brazil

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

Solution

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

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