I just tried to prove the very elementary one, the one that you typed. I hope the proof is correct. PS: the wikipedia page doesn't give any hint of the proof.
For a function $f\in C^2(-\infty, \infty)$, we have the following Taylor expansion:
$$ f(t)=f(r)+(t-r)f'(r)+\int_r^t (t-s)f''(s)ds$$
So
$$ f'(r)=\frac{f(t)-f(r)}{t-r}-\frac{1}{t-r}\int_r^t (t-s)f''(s)ds$$
Take absolute value on both sides,
$$ |f'(r)|\leq \left|\frac{f(t)-f(r)}{t-r}\right|+\frac{1}{|t-r|}\left|\int_r^t (t-s)f''(s)ds\right|\leq \frac{1}{|t-r|}|f(t)-f(r)|+\frac{|t-r|}{2}\|f''\|_\text{sup}\leq \frac{2\|f\|_\text{sup}}{|t-r|}+\frac{\|f''\|_\text{sup}}{2}|t-r|$$
Let $s=|t-r|$. Since we assume $f\in C^2(-\infty, \infty)$, so the above inequality is true for all $s$:
$$ |f'(r)|\leq \frac{1}{s}2\|f\|+\frac{\|f''\|}{2}s $$
So it's still true even we take the infimum over $s$:
$$ |f'(r)|\leq \inf_{s}\left(\frac{1}{s}2\|f\|+\frac{\|f''\|}{2}s\right) $$
Now let's discuss the infimum of the function
$$ g(s)=\frac{C}{s}+Ds $$
for some constants $C$ and $D$. Since $s$ can be taken from $-\infty$ to $\infty$, this function can achieve its infimum when $s=\sqrt{\frac{C}{D}}$ and its infimum is $2\sqrt{CD}$.
So for any $r$, we have
$$ |f'(r)|\leq 2\sqrt{2\|f\|_\text{sup}\times \frac{\|f''\|_\text{sup}}{2}}=2\sqrt{\|f\|\cdot \|f''\|} $$
Therefore, $\|f'\|_\text{sup}\leq 2\sqrt{ \|f\|\cdot \|f''\|}$ and take the square on both sides:
$$ \|f'\|^2\leq 4\|f\|\|f''\|$$
More general cases can see reference here:
http://ajmaa.org/RGMIA/papers/v6n2/inequalities.pdf
