# 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.

• Like to elaborate a bit why $\sqrt{(x^2+AD^2)(y^2+AD^2)}\geq xy+AD^2$ is suffficent?
– Mick
Commented Sep 16, 2016 at 17:19

we have $$h_a=\frac{2\sqrt{s(s-a)(s-b)(s-c)}}{a}$$ plugging this in the given inequality and simplifying and factorizing we get $$(b-c)^2 (-(a-b-c)) (a+b+c)\geq 0$$ and this is true.

Let $h_a=AD$

Heron's formula

$$\sqrt{s(s-a)(s-b)(s-c)}=\frac{1}{2}h a$$

where

$$s=\frac{a+b+c}{2}$$

written in terms of $a,b,c$ gives

$$\frac{(b+c)^2-a^2}{4h^2}=\frac{a^2}{a^2-(b-c)^2}$$

The right hand side is $\geq 1$ therefore

$$\frac{(b+c)^2-a^2}{4h^2}\geq 1$$

so

$$(b+c)^2\geq a^2+ 4h^2$$

Equality holds if and only if $b=c$

• Please tell me why the right hand side is greater than 1 Commented Sep 16, 2016 at 14:28
• @sufaidsaleel The numerator is obviously greater than or equal to the denominator $a^2\geq a^2-(b-c)^2\geq 0$ Commented Sep 16, 2016 at 14:48