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Prove that if $a,b,$ and $c$ are positive real numbers then $$(a+b)(a+c) \geq 2 \sqrt{abc(a+b+c)}.$$

This looks like a simple question. We can apply AM-GM twice to get $(a+b)(a+c) \geq 4a\sqrt{bc}$. Then how do I use that fact to get $(a+b)(a+c) \geq 2 \sqrt{abc(a+b+c)}$?

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We have $$(a+b)(a+c) = a(a+b+c) +bc.$$ We can now apply AM-GM.

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    $\begingroup$ You mean AM-GM. $\endgroup$ – user236182 Jan 10 '16 at 2:18
  • $\begingroup$ Yes, AM-GM inequality. $\endgroup$ – nguyen0610 Jan 10 '16 at 2:44
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we get after squaring and factorizing $$ \left( {a}^{2}+ab+ac-cb \right) ^{2}$$ and this is nonnegative

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  • $\begingroup$ The point is to show how you got to the result. Brutal force surely would not be a good choice. $\endgroup$ – cr001 Jan 10 '16 at 7:18

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