Inequality involving two altitudes of an isosceles triangle and its base I am trying to solve the following multiple choice problem:
$ABC$ is a triangle such that $AB=AC$. Let $D$ be the foot of the perpendicular from $C$ to $AB$ and $E$ the foot of the perpendicular from $B$ to $AC$. Then


*

*$BC^3<BD^3+BE^3$

*$BC^3=BD^3+BE^3$

*$BC^3>BD^3+BE^3$

*none of the foregoing statements need always be true.


I have tried the following steps:

Since $BC$ is the hypotenuse for the right angled triangles $BCD$ and $BCE$, so $BC>BE$ and $BC>BD$, so that $BC^3>BE^3$ and $BC^3>BD^3$. Thus, $BC^3>\dfrac{BD^3+BE^3}{2}$. But the answer given in my textbook is option $3$ i.e. $BC^3>BD^3+BE^3$. But I can't get it. i need some help in this regard.

By symmetry we have $BD=CE$. Now from the $BCE$ right triangle we have $$BC^2=CE^2+BE^2=BD^2+BE^2$$
Hence $BC^3=BC\times BD^2+BC\times BE^2>BD^3+BE^3$ (because $BC>BD$ and $BC>BE$).
