Why is $X = \{x:3x_{1}^{2}-2x_{2}^{2}\geq1, x_{1},x_{2}\geq 0\}$ a convex set

Question

The question is to determine if the set $$X = \{x:3x_{1}^{2}-2x_{2}^{2}\geq1, x_{1},x_{2}\geq 0\}$$ is convex.

I have a theorem which states that if $g(x)$ is a convex function then $g(x)\leq b$ also determines a convex function. This makes me want to investigate $$f(x_{1},x_{2})=2x_{2}^{2}-3x_{1}^{2}.$$ Because if $f$ is convex so is $2x_{2}^{2}-3x_{1}^{2}\leq1$ and multiplying both sides by $-1$ we get the original expression.

Another theorem we have is that a twice differentiable function $f$ is convex if the Hessian matrix of $f$, denoted $H_{f}(x)$, is a positive semidefinite matrix (I.e has non-negative eigenvalues).

So, setting up the Hessian for $f$ I get that $$H_{f}(x)=\begin{pmatrix}-6 & 0\\ 0 & 4\\\end{pmatrix}$$ clearly the eigenvalues are $-6$ and $4$ , but then the Hessian is not positive semidefinite implying the function is not convex.

Question: The solutions manual says that $X$ is a convex set, is that correct? and if so, what mistakes have I made?

Answer 1

$$X = \{(x,y)\in \mathbb{R}^2_+ \text{ }|\text{ }3y^{2}-2x^{2}\geq1\} = \lbrace(x,y)\in \mathbb{R}^2_+ \text{ }|\text{ }y \geq \sqrt{\frac{2x^{2}+1}{3}} \rbrace $$

so by convexity of $x \mapsto \sqrt{\frac{2x^{2}+1}{3}} $, the set $X$ is convex.