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 ?
 A: 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    $  
A: Let's suppose the triangle is acute  

We have
$$
a_1^2 + h^2 = b^2\\
a_2^2 + h^2 = c^2
$$
and by Cauchy-Schwarz inequality
$$
a_1\cdot a_2 + h\cdot h \leq \sqrt{a_1^2 + h^2}\sqrt{a_2^2 + h^2} = bc
$$
Summing the above relations we get
$$
a^2 + 4h^2 = a_1^2 + a_2^2 + 2a_1 a_2 + 4h^2 \leq b^2 + c^2 +2bc = (b + c)^2
$$
A similar reasoning shows the inequality is true even for obtuse triangles.
A: 

For acute/right angled triangle $h=b\sin C=c\sin B$  and 
for obtuse triangle (let $\angle B>\frac \pi 2$), $h=c\sin(\pi-B)=c\sin B$
$4h^2=2h\cdot 2h=(2b\sin C)(2c\sin B)$
$=2bc(\cos(B-C)-\cos(B+C))$
$=2bc(\cos(B-C)+\cos A)$ 
as $\cos(B+C)=\cos(\pi-A)=-\cos A$
We know, $$\cos A=\frac{b^2+c^2-a^2}{2bc}$$
$$\implies 1+\cos A=\frac{b^2+c^2-a^2}{2bc}+1=\frac{(b+c)^2-a^2}{2bc}$$
$\implies (b+c)^2-a^2=2bc (1+\cos A)$
$4h^2-((b+c)^2-a^2)$
$=2bc(\cos(B-C)+\cos A)-2bc (1+\cos A)$
$=2bc(\cos(B-C)-1)\le0$ as $\cos(B-C)\le 1$ and $bc>0$
