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 ?



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 $bc>0$
