Prove that the ellipsoid $E = \{ (x,y,z) \in R^3 \mid \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \leq 1 \}$ is convex I know what an ellipsoid is, but I do not know how to approach how to prove that one is convex. Any advice would be greatly appreciated.
 A: Let $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ be two points in the ellipsoid. We have to prove that $(\lambda x_1+(1-\lambda)x_2,\lambda y_1+(1-\lambda)y_2,\lambda z_1+(1-\lambda)z_2)$ also lies in the ellipsoid, where $0 \le \lambda \le 1$.
Let $$A = \frac{\left(\lambda x_1+(1-\lambda)x_2\right)^2}{a^2}+\frac{\left(\lambda y_1+(1-\lambda)y_2\right)^2}{b^2}+\frac{\left(\lambda z_1+(1-\lambda)z_2\right)^2}{c^2}.$$
It is easy to observe that
$$A = \lambda^2\left(\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} + \frac{z_1^2}{c^2}\right) + (1-\lambda)^2\left(\frac{x_2^2}{a^2} + \frac{y_2^2}{b^2} + \frac{z_2^2}{c^2}\right)+2\lambda(1-\lambda)\left(\frac{x_1x_2}{a^2} + \frac{y_1y_2}{b^2} + \frac{z_1z_2}{c^2}\right).$$
Consequently,
$$A \le \lambda^2 + (1-\lambda)^2+2\lambda(1-\lambda)\left(\frac{x_1x_2}{a^2} + \frac{y_1y_2}{b^2} + \frac{z_1z_2}{c^2}\right).$$
By Cauchy-Schwarz inequality,
$$\left(\frac{x_1x_2}{a^2} + \frac{y_1y_2}{b^2} + \frac{z_1z_2}{c^2}\right) \le \sqrt{\left(\frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} + \frac{z_1^2}{c^2}\right)\cdot \left(\frac{x_2^2}{a^2} + \frac{y_2^2}{b^2} + \frac{z_2^2}{c^2}\right)} \le 1.$$
Therefore,
$$A \le \lambda^2 + (1-\lambda)^2+2\lambda(1-\lambda)=1.$$
Q.E.D.
