Prove that $(b+c)^2\geq a^2+4\cdot AD^2$ 
Let $ABC$ be a triangle and $AD$ be the altitude through $A$. Prove that $$(b+c)^2\geq a^2+4\cdot AD^2$$
  (where $a=BC$, $b=CA$, $c=AB$).

I used Apollonius theorem and Pythagoras theorem every where. I guess that we can do it using these two theorems but I can't process.

Let $h_a=AD$
Heron's formula
$$\sqrt{s(s-a)(s-b)(s-c)}=\frac{1}{2}h a$$
where 
$$s=\frac{a+b+c}{2}$$
written in terms of $a,b,c$ gives
$$\frac{(b+c)^2-a^2}{4h^2}=\frac{a^2}{a^2-(b-c)^2}$$
The right hand side is $\geq 1$ therefore
$$\frac{(b+c)^2-a^2}{4h^2}\geq 1$$
so
$$(b+c)^2\geq  a^2+ 4h^2$$
Equality holds if and only if $b=c$ 
