It is well known (see Upper bound for the sum of absolute values of the eigenvalues) that for a matrix $A=(a_{ij})\in\mathbb{R}^{n\times n}$ $$ \sum_{i=1}^n s_i(A)\leq\sum_{i,j=1}^n |a_{ij}|, $$ where $s_i(A)$ denote the singular values of $A$. I am now wondering whether the following generalization $$ \sum_{i=1}^n \sqrt{s_i(A)}\leq\sum_{i,j=1}^n \sqrt{|a_{ij}|} $$ still holds true.
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In fact, the assertion is true. To see this let $A=U|A|$ be a polar decomposition of $A$ and denote its $i$-th column by $a_i$. Then $$ \sum_{i=1}^n \sqrt{s_i(A)}=\operatorname{tr}(|A|^{1/2})=\operatorname{tr}[(U^TA)^{1/2}]=\sum_{i=1}^n (U^TA)^{1/2}(i,i)\leq \sum_{i=1}^n \sqrt{(U^TA)(i,i)}, $$ where the last step used Peierls-Jensen's inequality. Since $U$ is orthogonal, we infer $$ \sum_{i=1}^n \sqrt{(U^TA)(i,i)}\leq\sum_{i=1}^n\sqrt{\|Ua_i\|_2}=\sum_{i=1}^n\sqrt{\|a_i\|_2}\leq \sum_{i=1}^n\sqrt{\|a_i\|_1}. $$ The proof is finished by using the elementary inequality $\sqrt{a+b}\leq\sqrt{a}+\sqrt{b}$, $a,b\geq 0$, for $n-1$ times.
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