From looking at the images you posted it seems that you are proving:
$\displaystyle |V\cdot U|\le\|V\|\|U\|$
The standard proof follows from the fact $\displaystyle \frac{V\cdot U}{\|V\|\|U\|}=\cos(\theta)$ where $\displaystyle \theta$ is the angle between the two non-zero vectors $\displaystyle V~\&~U$.
Then note that $\displaystyle 0\le |\cos(\theta)|\le 1~.$
first of all, you have $\displaystyle \|U\|$ and $\displaystyle \|V\|$ wrong.
$\displaystyle \|U\| = \sqrt{U \cdot U} = \sqrt{\int_0^{\frac{\pi}{2}} \sin^2x\ dx}$
$\displaystyle = \sqrt{\int_0^{\frac{\pi}{2}}(1 - \cos^2x)\ dx}$
$\displaystyle = \sqrt{\frac{\pi}{2} - \int_0^{\frac{\pi}{2}} \cos^2x\ dx}$
$\displaystyle = \sqrt{\frac{\pi}{2} - \int_0^{\frac{\pi}{2}} \frac{1 + \cos 2x}{2}\ dx}$
$\displaystyle = \sqrt{\frac{\pi}{2} - \frac{\pi}{4} - \frac{1}{4}\int_0^{\frac{\pi}{2}} \cos 2x\ 2dx}$
$\displaystyle = \sqrt{\frac{\pi}{4} - \frac{1}{4}(\sin (\pi) - \sin (0))}$
$\displaystyle = \frac{\sqrt{\pi}}{2}$
while:
$\displaystyle \|V\| = \sqrt{V \cdot V} = \sqrt{\int_0^{\frac{\pi}{2}} \cos^2x\ dx}$
$\displaystyle = \sqrt{\int_0^{\frac{\pi}{2}} \frac{1+\cos 2x}{2}\ dx} $
$\displaystyle = \sqrt{\frac{\pi}{4} + \frac{1}{4}(\sin(\pi) - \sin(0))}$
$\displaystyle = \frac{\sqrt{\pi}}{2}$, as well.
thus:
$\displaystyle \|U\|\|V\| = \frac{\pi}{4}$.
on the other side of the inequality we have:
$\displaystyle |U \cdot V| = \left|\int_0^{\frac{\pi}{2}} \sin x \cos x\ dx \right| $
$\displaystyle = \left| \frac{\sin^2(\frac{\pi}{2})}{2} - \frac{\sin^2(0)}{2} \right| = \left|\frac{1}{2} \right| = \frac{1}{2} $
since $\displaystyle 1 < \frac{\pi}{2} $, we indeed have for these particular U,V:
$\displaystyle |U \cdot V| \leq \|U\|\|V\|$.
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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!
still incorrect. for starters, you have an extra 2^{-n} term in $\displaystyle \|U\|$ and $\displaystyle \|V\|$.
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:
$\displaystyle \|U\| = \sqrt{\sum_{n=1}^{\infty} \frac{1}{2^n}} = 1$
and for V, you should have:
$\displaystyle \|V\| = \sqrt{\sum_{n=1}^{\infty} \frac{1}{8^n}} = \frac{1}{\sqrt{7}}$
while for the dot product, you should have:
$\displaystyle |U \cdot V| = \left| \sum_{n=1}^{\infty} \frac{1}{4^n} \right| = \frac{1}{3}$