Sides of triangle and an altitude Let $a$, $b$, $c$ be the lengths of the sides of a triangle. Let $h$ be the altitude drawn on the side of length $a$   Then is  $a^2 + 4h^2 - (b+c)^2$ always negative ?

I also have been able to come up with a proof that the expression indeed is negative ;
$-(b-c)^2 ≤ 0 $
$\implies a^2 - (b-c)^2 ≤ a^2 $
$\implies (a+b-c)(a-b+c) ≤ a^2 $
$\implies  (a+b+c)(b+c-a)(a+b-c)(a-b+c) ≤ a^2 ( a+b+c)(b+c-a)  $
 {since $a+b+c >0$ and for a triangle $b+c > a$ , the multiplication in the above line doesn't change sign of inequality} 
$\implies  16 A^2 ≤ a^2 \{ (b+c)^2 - a^2 \} $ [ A is the area of the triangle , by Heron's formula ] 
$\implies 16 (  \frac 12 ah )^2 ≤ a^2 \{(b+c)^2 - a^2\}  $  [ since h is altitude on a , A= ah/2 ]
$\implies  4 (ah)^2  ≤ a^2 \{ (b+c)^2 - a^2 \} $
$\implies  4 h^2  ≤ (b+c)^2 - a^2    $  
