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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 ?

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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 $bc>0$

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That's a wonderful proof ! I have also given a proof , don't forget to see it –  Souvik Dey Oct 19 '12 at 10:51

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 $

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Let's suppose the triangle is acute

triangle

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.

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