If $a,b,c$ are positive reals, then prove that $(a+b)(a+c)\ge 2 \sqrt {abc(a+b+c)}$

Question

If $a,b,c$ are positive reals, then prove that $(a+b)(a+c)\ge 2 \sqrt {abc(a+b+c)}$

My tries:
By applying AM-GM inequality,
$$a+b\ge 2\sqrt {ab}$$ and, $$a+c\ge2\sqrt{ac}$$ and clearly LHS $=a^2+ac+ba+bc$, which is $\ge4a\sqrt{bc}$
Similarly, $(b+a)(b+c)\ge4b\sqrt{ac}$ and $(c+a)(c+b)\ge4c\sqrt{ab}$
What next?(I gave the Cauchy Schwarz tag because I do not know if this can be solved by Cauchy Schwarz inequality.)

Answer 1

You have $$a(a+b+c) + bc \geq 2\sqrt{abc(a+b+c)}.$$