<s>[INST] Showing that a constraint is not convex 
Is the following problem convex?
$$\begin{array}{ll} \text{minimize} & x+y^2\\ \text{subject to} & x + y \leq 2\\ & \frac{x+y^2}{x^2+2} \leq 3\\ & x^2 = 1\end{array}$$

The problem is, in fact, not convex, but I'm having problems showing that the constraint $$\frac{x+y^2}{x^2+2} \leq3$$ is not convex. Is there a way to do so without calculating the Hessian?
 [/INST] $$\frac{x+y^2}{x^2+2}\le 3$$
$$y^2 \le 3x^2-x+6$$
Hence if $y$ is fixed, and $x$ is large enough in magnitude, then this construct will be satisfied. 
Let $y=3$, then $(2,3)$ and $(-2,3)$ both satisfies the inequality, but $(0,3)$ doesn't.
Remark:
For this question, we know that $x=1$ or $x=-1$. 
If $x=1$, $y \le 1$ and $y^2 \le 8$, hence $-\sqrt{8} \le y \le 1.$ 
If $x=-1$, $y \le 3$ and $y^2 \le 10$, hence $-\sqrt{10} \le y \le 3.$
In particular $(1,0)$ and $(-1,0)$ are both feasible but $(0,0)$ is not feasible.
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