I include some information about Hölder's inequality just for completion of details for Mikael's nice [answer](https://mathoverflow.net/a/78331).

The Schatten-$p$ norm of a matrix $X$ is defined as
$$\lVert X\rVert_p := \Bigl(\sum\nolimits_i \sigma_i^p(X)\Bigr)^{1/p},$$
where $\sigma_i(X)$ denotes the $i$-th singular value of $X$.

From this definition, we see that $\lVert X\rVert_p = \lVert\sigma(X)\rVert_p$, where $\sigma(X)$ is the *vector* of singular values. 

From [von Neumann's trace inequality][1] we know that
$$|{\operatorname{trace}(XY)}| \le \langle \sigma(X), \sigma(Y)\rangle,$$

while from the Hölder inequality for vectors, we have
$$\langle \sigma(X), \sigma(Y)\rangle \le \lVert\sigma(X)\rVert_p\lVert\sigma(Y)\rVert_q,$$
where $1/p + 1/q = 1$.

On combining the above two, we immediately get the alleged Schatten-$p$ norm Hölder inequality.


  [1]: https://en.wikipedia.org/wiki/Trace_inequality#Von_Neumann%27s_trace_inequality_and_related_results