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anonymous
 3 years ago
@mukushla
Prove \(ab + bc + ac \ge \sqrt{3abc \left(a+b+c\right)}\).
anonymous
 3 years ago
@mukushla Prove \(ab + bc + ac \ge \sqrt{3abc \left(a+b+c\right)}\).

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0could u plz typr it again?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i am sorry, i redid it. refresh.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it just needed squaring and then it becomes, \[a^2b^2 + c^2b^2 + a^2c^2 \ge abc(a+b+c)\] and then a simple complete square. i act so foolishly sometimes, sorry for troubling you mukushla

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0squaring both sides \[a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)≥ 3abc(a+b+c)\\ a^2b^2+b^2c^2+c^2a^2abc(a+b+c)\ge 0 \\2a^2b^2+2b^2c^2+2c^2a^22abc(a+b+c)\ge 0\\a^2b^22a^2bc+a^2c^2+a^2b^22ab^2c+b^2c^2+a^2c^22abc^2+b^2c^2\ge0\\ (ab−bc)^2+(bc−ca)^2+(ca−ab)^2\ge0\]
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