# Does Cauchy-Schwarz hold for: $\langle\textbf{u},\textbf{v}\rangle \;\leq ||\textbf{u}|| \cdot ||\textbf{v}||$

I am wondering whether the Cauchy-Schwarz inequality does hold when absolute value is not considered for the LHS.

Let me explain: In standard Cauchy-Schwarz we have:

$$| \langle \textbf{u},\textbf{v}\rangle |\;\leq \|\textbf{u}\| \cdot \|\textbf{v}\|$$

But will this hold if no absolute value is taken for the left hand side?

My thoughts:

Since $$| \langle\textbf{u},\textbf{v}\rangle |\; = \|\textbf{u}\| \cdot \|\textbf{v}\| \cos(\alpha)$$

we have that since $$\cos(\alpha) \leq 1$$, then the inequality:

$$\langle\textbf{u},\textbf{v}\rangle \;\leq \|\textbf{u}\| \cdot \|\textbf{v}\|$$, should hold.

• Are you allowing complex values for $\left<u,v\right>$? – Angina Seng Mar 1 '19 at 6:55
• Thanks for your reply. Not really, I assume $u$ and $v$ are real. (by the way, what is the $\LaTeX$ sintax for inner product? – kentropy Mar 1 '19 at 6:56
Note that $$\langle u,v\rangle\leqslant\bigl\lvert\langle u,v\rangle\bigr\rvert\leqslant\lVert u\rVert.\lVert v\rVert$$.