Where am I wrong in my understanding about the activeness of the constraints? I have following convex optimization problem $$\text{min.      }~~ x~\\ \text{s.t.}~~~\frac{y^2}{x}\leq z\\ y+z\leq c$$ where $\{x,y,z\}$ are the non-negative variables and $c$ is some positive constant. I think in the optimal solution both of the constraints should be active due to following reasoning.
1- Suppose that for the optimal solution we have $\{x^*,y^*,z^*\}$ and we consider the following possibilities 
1a- First constraint is not active. Then we can decrease $x$ (hence reducing the objective function) to make the constraint active and hence $\{x^*,y^*,z^*\}$ is not the optimal solution.
1b- Second constraint is not active. In this case, we can increase $z$ to make the right hand side of first constraint bigger and subsequently we can reduce $x$ to reduce the objective function.
Based on above reasoning, I think, both of the constraints should be active. However, when I solve the above problem through cvx in MATLAB I see that the constraints are not active. Where am I wrong in my understanding. Thanks in advance. 
 A: Your reasoning is right. However, you have overseen that you can choose $y$ arbitrarily small when $x$ approaches 0. Therefore, the constraints will not be met with equality.
For example, you may choose $z=c/2, y = x/2$. Then assume $x$ goes towards 0. 


*

*The second constraint will be met as $y + z = x/2 + c/2 < c$ when $x$ is small enough

*The first constrain will be met as $\frac{y^2}{x} = x/4 < z = c/2$ when $x$ is small enough.


Consequently, we have the solution $x^\star \rightarrow 0 $ of the optimization problem. 
A: With the help of some slack variables $\epsilon_i$ we can build the lagrangian
$$
L(x,y,z,\lambda,\epsilon) = x-\lambda_1\left(\frac{y^2}{x}-z+\epsilon_1^2\right)+\lambda_2(y+z-c+\epsilon_2^2) + \lambda_3(x-\epsilon_3^2)+\lambda_4(y-\epsilon_4^2)+\lambda_5(z-\epsilon_5^2)
$$
solving the stationary points for
$$
\nabla L = 0
$$
we obtain the solution
$$
\left[
\begin{array}{ccccccccc}
x&y&z&\epsilon_1^2&\epsilon_2^2&\epsilon_3^2&\epsilon_4^2&\epsilon_5^2& x\\
-4c&2c&-c&0&0&-4c&2c&-c&-4c\\
\end{array}
\right]
$$
which shows us that the only feasible solution is for $c = 0$ because $\epsilon_i^2 \ge 0$
