I include some information about Hölder's inequality just for completion of details for Mikael's nice answer.

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

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

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

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

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


  [1]: http://en.wikipedia.org/wiki/Von_Neumann%27s_trace_inequality