From looking at the images you posted it seems that you are proving:
$|V\cdot U|\le\|V\|\|U\|$

The standard proof follows from the fact $\frac{V\cdot U}{\|V\|\|U\|}=\cos(\theta)$ where $\theta$ is the angle between the two non-zero vectors $V~\&~U$.

Then note that $0\le |\cos(\theta)|\le 1~.$

first of all, you have $\|U\|$ and $\|V\|$ wrong.

$\|U\| = \sqrt{U \cdot U} = \sqrt{\int_0^{\frac{\pi}{2}} \sin^2x\ dx}$

$= \sqrt{\int_0^{\frac{\pi}{2}}(1 - \cos^2x)\ dx}$

$= \sqrt{\frac{\pi}{2} - \int_0^{\frac{\pi}{2}} \cos^2x\ dx}$

$= \sqrt{\frac{\pi}{2} - \int_0^{\frac{\pi}{2}} \frac{1 + \cos 2x}{2}\ dx}$

$= \sqrt{\frac{\pi}{2} - \frac{\pi}{4} - \frac{1}{4}\int_0^{\frac{\pi}{2}} \cos 2x\ 2dx}$

$= \sqrt{\frac{\pi}{4} - \frac{1}{4}(\sin (\pi) - \sin (0))}$

$= \frac{\sqrt{\pi}}{2}$

while:

$\|V\| = \sqrt{V \cdot V} = \sqrt{\int_0^{\frac{\pi}{2}} \cos^2x\ dx}$

$= \sqrt{\int_0^{\frac{\pi}{2}} \frac{1+\cos 2x}{2}\ dx}$

$= \sqrt{\frac{\pi}{4} + \frac{1}{4}(\sin(\pi) - \sin(0))}$

$= \frac{\sqrt{\pi}}{2}$, as well.

thus:

$\|U\|\|V\| = \frac{\pi}{4}$.

on the other side of the inequality we have:

$|U \cdot V| = \left|\int_0^{\frac{\pi}{2}} \sin x \cos x\ dx \right|$

$= \left| \frac{\sin^2(\frac{\pi}{2})}{2} - \frac{\sin^2(0)}{2} \right| = \left|\frac{1}{2} \right| = \frac{1}{2}$

since $1 < \frac{\pi}{2}$, we indeed have for these particular U,V:

$|U \cdot V| \leq \|U\|\|V\|$.

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for your second problem, you are also calculating the norms of U and V incorrectly. use the definition of the inner product you are given!

Thank you so much!! Deveno! I missed the obvious point

still incorrect. for starters, you have an extra 2-n term in $\|U\|$ and $\|V\|$.

secondly, you don't seem to get that the inner product of two vectors should be a SCALAR, not an "expression".

so for U, you should have:

$\|U\| = \sqrt{\sum_{n=1}^{\infty} \frac{1}{2^n}} = 1$

and for V, you should have:

$\|V\| = \sqrt{\sum_{n=1}^{\infty} \frac{1}{8^n}} = \frac{1}{\sqrt{7}}$

while for the dot product, you should have:

$|U \cdot V| = \left| \sum_{n=1}^{\infty} \frac{1}{4^n} \right| = \frac{1}{3}$