Prove $\frac{( 4n+5 ) ( ( 2n+2 )! ) ^2}{2}( \int_{-1}^1{f( x ) \text{d}x} ) ^2\le \int_{-1}^1{( f^{( 2n+2 )}( x ) )^2 \text{d}x}$ Let $f \in C^{2n+2}[-1,1]$, and
$$
f\left( 0 \right) =f''\left( 0 \right) =\cdots =f^{\left( 2n+2 \right)}\left( 0 \right) =0
$$
Prove that
$$
\frac{\left( 4n+5 \right) \left( \left( 2n+2 \right) ! \right) ^2}{2}\left( \int_{-1}^1{f\left( x \right) \text{d}x} \right) ^2\le \int_{-1}^1{\left( f^{\left( 2n+2 \right)}\left( x \right) \right)^2 \text{d}x}
$$
I even don't know how to deal with the basic circumstances, such as when $n=0$, we have
$$
10\left( \int_{-1}^1{f\left( x \right) \text{d}x} \right) ^2\le \int_{-1}^1{\left( f''\left( x \right) \right) ^2\text{d}x}
$$
I tried to apply Cauchy–Schwarz inequality and Integration by parts, but they didn't work.
Also I think the general formula may relate to Taylor series since it has factorials on the left side.
Can anyone help?
 A: I will do the case $n = 0$ by considering @r9m's suggestion.
This means we have $f(0) = f''(0) = 0$ and also $g'(x) = f(x). g''(x) = f'(x)$ and $g'''(x) = f''(x).$ Therefore, expanding $g$ around the point $a = 0$ by Taylor's theorem:
$$g(x) = g(0) + g'(0)x + g''(0)\frac{x^2}{2} + \frac 12\int_0^x g'''(t)(x-t)^2dt = f'(0)\frac{x^2}{2} + \frac 12\int_0^xf''(t)(x-t)^2dt$$
Therefore,
$$\int_{-1}^1f(x)dx = g(1) - g(-1) = \frac 12\int_{0}^1f''(t)(1-t)^2dt+\frac 12\int_{-1}^0f''(t)(1+t)^2dt$$
and so the C-S on each term of the RHS gives:
$$\int_{0}^1f''(t)(1-t)^2dt\leq\left(\int_0^1 (f''(t))^2dt\right)^{1/2}\left(\int_0^1(1-t)^4dt\right)^{1/2} =\left(\int_0^1 (f''(t))^2dt\right)^{1/2}\dfrac{1}{\sqrt{5}} $$
and
$$\int_{-1}^0f''(t)(1+t)^2dt\leq\left(\int_{-1}^0 (f''(t))^2dt\right)^{1/2}\left(\int_{-1}^0(1+t)^4dt\right)^{1/2} =\left(\int_{-1}^0 (f''(t))^2dt\right)^{1/2}\dfrac{1}{\sqrt{5}}.$$
Finally then we have by another application of C-S:
$$\int_{-1}^1f(x)dx\leq\dfrac{1}{2\sqrt{5}}\left( \left(\int_0^1 (f''(t))^2dt\right)^{1/2} +\left(\int_{-1}^0 (f''(t))^2dt\right)^{1/2} \right)\leq\dfrac{1}{2\sqrt{5}}\sqrt{2\int_{-1}^1(f''(t))^2dt}$$
where the last inequality is just:
$$\sqrt{a}+\sqrt{b}\leq\sqrt{2(a+b)},$$
which is just Cauchy-Schwartz.
You can clearly see from here that this should be immediatly generalizable to any $n>0.$
Interesting to note here is that you will have equality when:
$$f''(x) = 
\begin{cases}
1-x,\quad 0\leq t\leq 1 \\
1+x,\quad -1\leq t\leq 0
\end{cases}$$
