Given $a_1$ and $a_2$ such that $a_1≥a_2^2≥1$, solve the following problem
\begin{array}{ll} \text{minimize} & \mbox{tr} X\\X\in\mathbb{R}^{2\times 2}\\\text{subject to} & \mathrm{det}X\geq a_1,\\&\frac{\mathrm{det}X}{\mathrm{det}X-X_{11}}(X_{22}-1)\geq a_2,\\&X>0.\end{array}
If we assume $X$ is diagonal, then $\mathrm{tr}X=2\sqrt{a_1}$. ($x_1=x_2=\sqrt{a_1}$). I wonder whether this is an optimal?
Help me please to simplify the problem. If you have an idea on how to reformulate this problem for CVX will be great as well.
Since $X$ is symmetric, we may write this as $$ X = \begin{pmatrix}a&b\\b&c\end{pmatrix} $$ we have $\operatorname{Tr}(A)=a+c$ and $\det A=ac-b^2$, so the optimization problem is of the form \begin{align} \min&\quad a+c\\ \mathrm{s.t.}&\quad ac-b^2\geqslant a_1\\ &\quad \frac{(ac-b^2)(c-1)}{ac-b^2-a}\geqslant a_2\\ &\quad a,b,c\geqslant 0. \end{align} I'm not sure how to solve this, but you may want to look into the Karush–Kuhn–Tucker conditions.