Find minimum of piecewise defined function I'm trying to prove that $(-1,0)$ is the unique solution of the convex optimization problem
$$\min_{(x,y)\in\mathbb{R}^2} x + 10 \max \{x^2+2y^2-1,0\}.$$
Unfortunately, I'm not sure how to approach this as the function to minimize contains the $\max$ function. I tried it by case distinction, i.e. assume $x^2+2y^2-1 \ge 0$. Then the function to minimize is just the polynomial $x + 10(x^2+2y^2-1)$. But the minimum of this polynomial $(-1/20,0)$ does not fulfill the condition from the case distinction $x^2+2y^2-1 \ge 0$ anymore.
Any help is highly appreciated.
 A: Case-$1$: $x^2+2y^2-1\leq0$. Problem simplifies to:
$$\
\min_{(x, y) \in \mathbb{R}^2} x \quad s.t. \quad x^2+2y^2-1\leq0
$$
It's easy to see that the minimum value is $-1$ and occurs at $(-1, 0)$. However, you can use Lagrange Multiplier to solve this as done for the case-2 below.
Case-$2$: $x^2+2y^2-1\geq0$. Problem simplifies to:
$$\
\min_{(x, y) \in \mathbb{R}^2} x+10(x^2+2y^2-1) \quad s.t. \quad x^2+2y^2-1\geq0
$$
Writing Lagrangian for the problem:
$$\
L(x, y, \lambda)=x+10(x^2+2y^2-1) - \lambda(x^2+2y^2-1) \quad;\lambda\geq0
$$
Setting $\frac{\partial L}{\partial x}=0$ and $\frac{\partial L}{\partial y}=0$ we get,
$$\
x=\frac{1}{2\lambda-20}\text{ and } y=0
$$
Substituting it back in the constraint, we get $\lambda=\frac{19}{2}$ or $\lambda=\frac{21}{2}$. Finally, you can plug it back and verify that $\lambda=\frac{19}{2}$ corresponds to the global minimum giving us $(-1, 0)$ as the solution.  
In case if you are wondering what happens to $\lambda=\frac{21}{2}$. Its a local maximum.
Note: You can argue directly that if constraint is inactive i.e. $x^2+2y^2-1>0$. There exists no optimal for the objective and if the constraint is active, case-2 and case-1 are exactly same.
A: Where $x^2+2y^2-1\le0$, the objective function reduces to $x$. The smallest value of $x$ that satisfies $x^2+2y^2-1=0$ is $x=-1$, when $y=0$.
