linear algebra - Different version of Cauchy-Schwarz Inequality?

Show that

$$\|AC\|_2 \leq \|A\|_2 \|C\|_2,$$

where

$$A, C$$ are square matrices of the same size.

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I understand this probably needs to use Cauchy-Schwarz Inequality, but I don't know how should I start the problem, any help?

Answer

Observe that

$$\|AC\|=\sup_{\|x\|=1}\|ACx\|\le \sup_{\|x\|=1}\|A\|\|Cx\|=\|A\|\sup_{\|x\|=1}\|Cx\|=\|A\|\|C\|.$$
Here we only used that
$$\|ACx\|\le \|A\|\|Cx\|$$

which is an implication of the definition of $\|A\|$.

Note. This inequality HAS NOTHING TO DO with Cauchy-Schwarz, although it looks the same!