Prove that the ellipsoid \begin{align*} E = \left\{(x,y,z)\in\textbf{R}^{3}\mid\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} \leq 1\right\} \end{align*}
is convex.
MY ATTEMPT
To start with, let us consider the inner product defined by \begin{align*} \langle\textbf{x},\textbf{y}\rangle = \frac{x_{1}y_{1}}{a^{2}} + \frac{x_{2}y_{2}}{b^{2}} + \frac{x_{3}y_{3}}{c^{2}} \end{align*}
It is not hard to prove such function is bilinear, symmetric and positive-definite.
Thus $E$ is the unit ball accordingly to such inner product. Hence, if we take two points $\textbf{x}$ and $\textbf{y}$ in $E$, its convex combination also belongs to $E$. Indeed, one has \begin{align*} \langle t\textbf{x} + (1-t)\textbf{y},t\textbf{x} + (1-t)\textbf{y}\rangle & = t^{2}\|\textbf{x}\|^{2} + 2t(1-t)\langle\textbf{x},\textbf{y}\rangle + (1-t)^{2}\|\textbf{y}\|^{2}\\\\ & \leq t^{2}\|\textbf{x}\|^{2} + 2t(1-t)|\langle\textbf{x},\textbf{y}\rangle| + (1-t)^{2}\|\textbf{y}\|^{2}\\\\ & \leq t^{2}\|\textbf{x}\|^{2} + 2t(1-t)\|\textbf{x}\|\|\textbf{y}\| + (1-t)^{2}\|\textbf{y}\|^{2}\\\\ & \leq t^{2} + 2t(1-t) + (1-t)^{2} = 1 \end{align*}
Could someone double-check my solution? Any comment is welcome.
Alternative approach using norm:
Let $A=\operatorname{diag}(\frac1{a}, \frac1{b}, \frac1{c})$.
If $x^TA^2x=\|Ax\|^2 \le 1 $ and $y^TA^2y=\|Ay\|^2 \le 1 $ $\iff \|Ax\| \le 1$ and $\|Ay\|\le1$,
Let $t \in (0,1)$, $$\|A(tx+(1-t)y)\| \le t\|Ax\| + (1-t) \|Ay\| \le t + (1-t)=1$$
where I have used the triangle inequality.