I was reading some Linear Algebra contents and then I encountered this problem, I tried to prove this by using for example Holder inequality for Schatten norms, But I didn't succeed. The question is :
Show that for any Hermitian operators $M, \sigma \in \operatorname{Herm}(A),$ with $\sigma \geqslant 0,$ the following holds: $$ \|\sqrt{\sigma} M \sqrt{\sigma}\|_{1} \leqslant\|M\|_{2}\|\sigma\|_{2} $$
when dealing with Schatten norms, which explicitly use singular values $\sigma_k$, it's a really bad idea to overload this and use $\sigma$ to denote a matrix, so I restate the problem as follows
for HPSD $B$, and Hermitian $M$ prove:
$\|B^\frac{1}{2} M B^\frac{1}{2} \|_{S_1} \leq\|M\|_{2}\|B\|_{S_2}$
Let $\Sigma_X$ denote a diagonal matrix containing the singular values of $X$ in order from largest to smallest.
proof 1: (less machinery)
$\|B^\frac{1}{2} M B^\frac{1}{2} \|_{S_1}$
$\leq \| M B \|_{S_1}$
$=\text{trace}\Big(U^*M B\Big)$
$\leq \text{trace}\Big(\Sigma_{U^*M}\Sigma_{B}\Big)$
$= \text{trace}\Big(\Sigma_{M}\Sigma_{B}\Big)$
$\leq \text{trace}\Big(\big(\Sigma_{M}\big)^2\Big)^\frac{1}{2}\cdot \text{trace}\Big(\big(\Sigma_{B}\big)^2\Big)^\frac{1}{2}$
$= \Big\Vert M\Big\Vert_{S_2}\cdot \Big\Vert B\Big\Vert_{S_2}$
where the inequalities are
(i) $B^\frac{1}{2} M B^\frac{1}{2}$ is Hermitian, hence Normal, so if we apply Schur Triangularization it is diagonal, with singular values given by the eigenvalues (and rescaled by a point on the unit circle as needed). On the other hand $M B$ has the same eigenvalues but in general is not normal, so it has a higher Schatten 1 norm. To make this explicit, where $MB$ may be Schur triangularizated by $Q$ and with some optimally chosen unitary diagonal matrix $D$
$\sum_{k=1}^n \big\vert\lambda_k^{(MB)}\big\vert =\sum_{k=1}^n \big\vert\lambda_k^{(B^\frac{1}{2}MB^\frac{1}{2})}\big\vert =\big\Vert B^\frac{1}{2} M B^\frac{1}{2}\big\Vert_{S_1}$ but by von Neumann Trace Inequality
$\sum_{k=1}^n \big\vert\lambda_k^{(MB)}\big\vert =\text{trace}\Big(\big(QDQ^*\big) M B\Big)\leq \text{trace}\Big(\Sigma_{QDQ^*} \Sigma_{M B}\Big)=\big\Vert M B\big\Vert_{S_1}$
(ii) von Neumann Trace Inequality
(iii) Cauchy Schwarz
proof 2: (more general)
Now, by Polar decomposition we have $B^\frac{1}{2} M B^\frac{1}{2}=UP$
$\|B^\frac{1}{2} M B^\frac{1}{2} \|_{S_1}$
$=\text{trace}\Big(U^*B^\frac{1}{2} M B^\frac{1}{2}\Big)$
$\leq\text{trace}\Big(\Sigma_{U^*B^\frac{1}{2}} \Sigma_{M B^\frac{1}{2}}\Big)$
$=\text{trace}\Big(\Sigma_{B^\frac{1}{2}} \Sigma_{M B^\frac{1}{2}}\Big)$
$\leq\text{trace}\Big(\Sigma_{B^\frac{1}{2}} \Sigma_{M}\Sigma_{B^\frac{1}{2}}\Big)$
$=\text{trace}\Big(\Sigma_{M}\Sigma_{B^\frac{1}{2}}\Sigma_{B^\frac{1}{2}} \Big)$
$=\text{trace}\Big(\Sigma_{M}\Sigma_{B}\Big)$
$\leq \text{trace}\Big(\big(\Sigma_{M}\big)^2\Big)^\frac{1}{2}\cdot \text{trace}\Big(\big(\Sigma_{B}\big)^2\Big)^\frac{1}{2}$
$=\Big\Vert M\Big\Vert_{S_2}\cdot \Big\Vert B\Big\Vert_{S_2}$
where the inequalities are
(i) von Neumann Trace Inequality
(ii) the fact that $\Sigma_{XY} \preceq_w \Sigma_{X}\Sigma_{Y}$ with $\preceq_w$ denoting weak majorization
(iii) Cauchy-Schwarz
Notice that Hermicity of $M$ was not needed.