Prove V is a convex set

Question

The exercise is as below

Let $V = \{x \in \mathbb{R}^2: 2x_1^2+3x_2^2\leq4\}$ Prove: V is convex

I started the proof like this:

Let $x,y \in V$ be given. Let $\lambda \in [0,1]$. We want to show that $\lambda x + (1-\lambda) y \in V$.

Now, I know that I have to rewrite $2[\lambda x_1 + (1-\lambda)y_1]^2+3[\lambda x_2 + (1-\lambda)y_2]^2$ into something like

$\lambda(2x_1^2+3x_2^2) + (1-\lambda)(2y_1^2+3y_2^2) \leq \lambda*4+(1-\lambda)*4 = 4$

But I just can't get my head around it, can someone please make a step-by-step derivation? I am really hopeless at this point.

Answer 1

It is not so obvious how to proceed directly. A more abstract approach is useful: note that $F(x_1,x_2) = 2x_1^2+3x_2^2$ is convex (in fact, $d^2F/dx^2 = \begin{pmatrix}4&0\\ 0 & 6\end{pmatrix}$ is positive definite), so $$2[\lambda x_1 + (1-\lambda)y_1]^2+3[\lambda x_2 + (1-\lambda)y_2]^2 = F(\lambda x+(1-\lambda y)) \leq\\ \hspace{7cm} \lambda F(x) + (1-\lambda)F(y) \leq \lambda 4 + (1-\lambda)4 = 4$$