TITLE: Proving Holder's inequality for Schatten norms
QUESTION [10 upvotes]: Sticking to the finite dimensional case, Holder's inequality for Schatten norms is given by

$$\left\|AB\right\|_{S^1}\leq\left\|A\right\|_{S^p}\left\|B\right\|_{S^q}$$

for $A,B$ $n\times n$ matrices, $p,q\in[1,\infty]$, and $\frac{1}{p}+\frac{1}{q}=1$.
So using Young's inequality, the expression I have in mind is the following
\begin{align}
\frac{\left\|AB\right\|_{S^1}}{\left\|A\right\|_{S^p}\left\|B\right\|_{S^q}}=\frac{1}{\left\|A\right\|_{S^p}\left\|B\right\|_{S^q}}\sum_{i=1}^n|\sigma_i(AB)|&\overset{?}{\leq}\frac{1}{\left\|A\right\|_{S^p}\left\|B\right\|_{S^q}}\sum_{i=1}^n|\sigma_i(A)||\sigma_{\pi(i)}(B)|\\
&\overset{\text{YI}}{\leq}\frac{1}{p\left\|A\right\|_{S^p}^p}\sum_{i=1}^n|\sigma_i(A)|^p + \frac{1}{q\left\|A\right\|_{S^q}^q}\sum_{i=1}^n|\sigma_{\pi(i)}(B)|^q\\
&=\frac{1}{p} + \frac{1}{q}\\
&=1
\end{align}
Focusing in on the inequality under question,
$$\sum_{i=1}^n|\sigma_i(AB)|\overset{?}{\leq}\sum_{i=1}^n|\sigma_i(A)||\sigma_{\pi(i)}(B)|$$
we see that proving the Schatten version of Holder's inequality boils down to proving that there exists a permutation $\pi$ of the indices $\{1,...,n\}$ such that the above inequality holds.  Of course maybe this isn't true, but it's the hurdle I ran into when trying to adapt the standard proof of Holder's inequality to the Schatten case.
Also I don't strictly need all the absolute values since singular values are always non-negative, but originally I was considering only Hermitian matrices, so I decided to include them.
Edit:
After doing some numerical tests it looks like permuting the indices isn't necessary.  Thus to win the bounty I'm either looking for a proof that
$$\sum_{i=1}^n\sigma_i(AB)\leq\sum_{i=1}^n\sigma_i(A)\sigma_i(B)$$
or if you have some other proof of the Schatten Holder's inequality different than the one I've tried to adapt above, then that's fine too.

REPLY [3 votes]: An alternative proof which is based on more standard technology is as follows.  First observe that for any $A$ and $B$, there exists a unitary matrix $U$ such that
$$
||AB||_1=|\mathrm{Tr}((AU)^\dagger B)|.
$$
Indeed consider the singular value decomposition
$$AB=\sum_i \sigma_i(AB)a_i(AB)b_i(AB)^\dagger,$$
where the singular values $\sigma_i(AB)$ are positive and listed in decreasing order and $a_i$ and $b_i$ are orthonormal sets of vectors.  We may then take $U$ to be any unitary such that $U^\dagger a_i(AB)=b_i(AB)$ for all $i$.  Therefore, since $||AU||_p=||A||_p$ for any unitary $U$, the result will be proved if we can show that
$$ 
|\mathrm{Tr}(A^\dagger B)|\leq||A||_p||B||_q
$$
for all $A$ and $B$ (this inequality is also referred to as Holder's inequality for Schatten norms).  The proof of this inequality follows first by applying von Neumann's trace inequality, which says that
$$|\mathrm{Tr}(A^\dagger B)|\leq\sum_i\sigma_i(A)\sigma_i(B),$$
with the singular values again listed in decreasing order, and then the classical Holder's inequality for $L_p$ spaces, which says that for two complex vectors $n_i$ and $m_i$ we have
$$\sum_i|n_i||m_i|\leq (\sum_i |n_i|^p)^{1/p}(\sum_i|m_i|^q)^{1/q}.$$
The classical Holder inequality is proven using Young's inequality
$$ab\leq a^p/p+b^q/q,$$
which holds for all $a,b\geq 0$ and $p\geq 1$, $1/p+1/q=1$.  The proof of von Neumann's inequality is more involved, one relatively accessible proof by Mirsky based on doubly stochastic matrices (which has some relation to the proof given in the other answer here) is given in: https://link.springer.com/article/10.1007/BF01647331.