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Given $\Sigma=$diag$\{\sigma_1,\sigma_2,\cdots,\sigma_n\}$ and $a\in\mathbb{R}$. Find \begin{array}{ll}\min\limits_{{X:\,det(I+X)\geq a}} tr(X\Sigma).\end{array}

My attempt: If $X=$diag$\{x_1,x_2,\cdots,x_n\},$ then using AM-GM inequality \begin{equation}tr(X\Sigma)=\sum_{i=1}^nx_i\sigma_i=\sum_{i=1}^n(1+x_i)\sigma_i-\sum_{i=1}^n\sigma_i\geq\frac{1}{n}\sqrt[n]{\prod_{i=1}^n(1+x_i)\sigma_i}-\sum_{i=1}^n\sigma_i\geq\frac{a}{n}\sqrt[n]{\prod_{i=1}^n\sigma_i}-\sum_{i=1}^n\sigma_i.\end{equation}

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Your computation contains $2$ mistakes; moreover, your problem is ill posed.

i) We assume that, for every $i$, $\sigma_i\geq 0$ and $\sigma_1>0$.

Note that $tr(X\Sigma)=\sum_ix_{i,i}\sigma_i$ where $X=[x_{i,j}]$. With the sole condition on $X$, $\det(I+X)\geq a$, the lower bound of $tr(X\Sigma)$ is $-\infty$; indeed, choose $X_t=diag(-tI_{2k},0_{n-2k})$ where $t\rightarrow +\infty$. Thus, we must add a condition on $X$; for example, its eigenvalues are real and $\geq -1$.

ii) We assume also that $a\geq 0$.

Let $f(X)=tr(X\Sigma)$ and $\phi(X)=\log(\det(I+X))-\log(a)$. We calculate the critical poins of the problem: "find $\min(f(X))$ under the condition $\phi(X)=0$".

$Df_X-\lambda D\phi_X:H\rightarrow tr(H\Sigma)-\lambda tr(H(I+X)^{-1})$. The critical points $X$ satisfy $\Sigma=\lambda(I+X)^{-1}$, that is, $I+X=\lambda\Sigma^{-1}$.

Thus $\det(I+X)=\lambda^n/\Pi_i \sigma_i=a$ and $\lambda=(\Pi_i\sigma_i)^{1/n}a^{1/n}$ (there is also the opposite solution when $n$ is even).

Finally, the candidate $X$ (to be a minimum) is a diagonal matrix: $X=(\Pi_i\sigma_i)^{1/n}a^{1/n}\Sigma^{-1}-I$. The associated value of $f$ is $tr(X\Sigma)=(\Pi_i\sigma_i)^{1/n}a^{1/n}n-\sum_i\sigma_i$.

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