Optimization of parallelepiped inside an ellipsoid Let $K \in R^3$  the ellipsoid given by the equation $ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 $ with $a,b,c > 0$ , let $(x,y,z) \in K$ on the first octant, consider the parallelepiped of vertices $(\pm x,\pm y,\pm z)$ inscribed on $K$  with volume $V = 8xyz$.
How can I find the maximum possible value of $V$?
I stuck with this hard problem for me i tried to find the explicit equation and then get the maximum values : Let $P=(x,y,z)$ be a point on the ellipsoid with $x,y,z\gt 0$.Then i took the eight different points with $P_i (\pm x,\pm y,\pm z)$  the vertices of a parallelepiped with the side length $2x , 2y$ and $2z$. 
Then, the volume parallelepiped is $V = 2x\cdot 2y\cdot 2z = 8 xyz$ and i remembered that $V$ is maximum if and only if $V^2$ is maximum .
Some help please.

Because
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$
We have 
$$\left(\frac{x^2}{a^2}\frac{y^2}{b^2}\frac{z^2}{c^2}\right)^{1/3} \le \frac{1}{3}\left(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}\right) = \frac{1}{3}$$
Thus
$$V=8|xyz| \le \frac{8abc}{\sqrt{27}}$$
