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

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 $(a+b)(a+c) \geq 2 \sqrt{abc(a+b+c)}$?

We have $$(a+b)(a+c) = a(a+b+c) +bc.$$ We can now apply AM-GM.
we get after squaring and factorizing $$\left( {a}^{2}+ab+ac-cb \right) ^{2}$$ and this is nonnegative