Integral inequality with a function twice differentiable Let $f:[0,1]\longrightarrow\mathbb{R}$ be a function twice differentiable with continous second derivative and $f(1)=f(0)$. The inequality:
$$\int_{0}^{1}(f''(x))^2dx\geq 120\left(\int_{0}^{1}xf'(x)dx\right)^2$$ holds?
 A: Let
$$
A=\int_0^1xf'(x)\,\mathrm{d}x\tag{1}
$$
Since $f(0)=f(1)$, we have
$$
\int_0^1f'(x)\,\mathrm{d}x=0\tag{2}
$$
$(1)$, $(2)$, and integration by parts gives
$$
\begin{align}
2A
&=\int_0^1(2x-1)f'(x)\,\mathrm{d}x\\
&=\int_0^1f'(x)\,\mathrm{d}x(x-1)\\
&=\int_0^1x(1-x)f''(x)\,\mathrm{d}x\tag{3}
\end{align}
$$
Apply Hölder to $(3)$:
$$
\begin{align}
4A^2
&\le\int_0^1[x(1-x)]^2\,\mathrm{d}x\int_0^1f''(x)^2\,\mathrm{d}x\\
&=\frac1{30}\int_0^1f''(x)^2\,\mathrm{d}x\tag{4}
\end{align}
$$
Plugging $(1)$ into $(4)$ yields
$$
120\left(\int_0^1xf'(x)\,\mathrm{d}x\right)^2\le\int_0^1f''(x)^2\,\mathrm{d}x\tag{5}
$$

Using $f(x)=x(1-x)(1+x(1-x))$, we see that $(5)$ is sharp: both sides equal $\dfrac{24}{5}$.
A: Since $f(1)=f(0)\Rightarrow \int_0^1f'(x)\,\mathrm{d}x=0$, By integration by parts,
\begin{equation}\label{MSE-730580}
\int_0^1g(x)f''(x)\mathrm{d}x=g(x)f'(x)\bigg|_0^1-\int_0^1g'(x)f'(x)\mathrm{d}x \tag{1}
\end{equation}
Let $g(x)$ be a polynomial defined as,
$$g(x)=ax^2+bx+c\Rightarrow g'(x)=2ax+b$$
Let
$$g(x)f'(x)\bigg|_0^1=0\Rightarrow g(0)=g(1)=0\Rightarrow b=-a$$
we can get,
$$g(x)=ax(x-1)\Rightarrow g'(x)=2ax-a \tag{2}$$
by $(1),(2)$, we have
$$\int_0^1x(x-1)f''(x)\mathrm{d}x=\int_0^1(1-2x)f'(x)\mathrm{d}x=-2\int_0^1xf'(x)\mathrm{d}x$$
by the Cauchy-Schwarz inequality,
\begin{align*}
\left(\int_0^1xf'(x)\mathrm{d}x\right)^2
&=\left(\int_0^1\Big(-\frac{1}{2}x(x-1)\Big)f''(x)\mathrm{d}x\right)^2\\
&\leqslant\int_0^1\Big(-\frac{1}{2}x(x-1)\Big)^2\mathrm{d}x\int_0^1(f''(x))^2\mathrm{d}x
\end{align*}
Therefore,
$$120\left(\int_0^1xf'(x)\,\mathrm{d}x\right)^2\le\int_0^1f''(x)^2\,\mathrm{d}x$$
because,
$$\int_0^1\bigg(-\frac{1}{2}x(x-1)\bigg)^2\mathrm{d}x=\frac{1}{120}$$
