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.

Suppose $DB=x$ and $DC=y$. Then $a\leq x+y$ (since $a=x+y$ or $a=|x-y|$), $b=\sqrt{x^2+AD^2},c=\sqrt{y^2+AD^2}$. It suffices to show $\sqrt{(x^2+AD^2)(y^2+AD^2)}\geq xy+AD^2$, which follows from Cauchy-Schwarz inequality.
