Proving $\int_0^1\vert(f(x)^2)'\vert dx\leq \int_0^1( f''(x))^2dx$, where $f(0)=f'(0)=0$ 
Suppose that $f\in C^{2}\left[ 0,1\right] $ and $f\left( 0\right) =f^{\prime
}\left( 0\right) =0$. Prove that
$$
\int_{0}^{1}\left\vert \left( f\left( x\right) ^{2}\right)
^{\prime }\right\vert dx\leq \int_{0}^{1} f^{\prime \prime }\left(
x\right)  ^{2}dx
$$

Using Cauchy-Schwarz inequality,
\begin{eqnarray*}
f^{\prime }\left( x\right)  &=&f^{\prime }\left( 0\right)
+\int_{0}^{x}f^{\prime \prime }\left( t\right) dt=\int_{0}^{x}f^{\prime
\prime }\left( t\right) dt \\
&\leq &\sqrt{x}\cdot\left[ \int_{0}^{x} f^{\prime \prime }\left( t\right)
 ^{2}dt\right] ^{1/2}
\end{eqnarray*}
So
$$
f^{\prime }\left( x\right) ^{2}\leq x\int_{0}^{x}
f^{\prime \prime }\left( t\right)^{2}dt\leq \int_{0}^{1}
f^{\prime \prime }\left( t\right) ^{2}dt,\text{ }\forall x\in \left[
0,1\right]
$$
and
$$
\int_{0}^{1} f^{\prime }\left( x\right) ^{2}dx\leq
\int_{0}^{1} f^{\prime \prime }\left( x\right) ^{2}dx.
$$
But I don't know how to get  $\int_{0}^{1}\left\vert \left( f\left( x\right)  ^{2}\right)
^{\prime }\right\vert dx $ in the left-hand member. Any help would be appreciated.
 A: First, by the Cauchy-Schwartz inequality, we have
$$|f(x)|=\big|f(0)+\int_0^x f'(t)\,dt\big| \leq \int_0^1 \Big|\text{1}_{[0,x]}(t)\cdot f'(t)\Big|\,dt \leq \sqrt{x}\cdot \Big(\int_0^1 (f')^2\Big)^{1/2}$$
Similarly, we have
$$|f'(x)| \leq \sqrt{x}\cdot \Big(\int_0^1 (f'')^2\Big)^{1/2}$$
Therefore
$$\big(\int_0^1 |(f^2(x))'| \,dx\big)^2=4\big(\int_0^1 |f\cdot f'|\big)^2\leq 4 \big(\int_0^1 f^2\big)\cdot \big(\int_0^1 (f')^2\big)$$
and
$$\int_0^1 (f')^2\leq \int_0^1 \big(x\cdot\int_0^1 (f'')^2\big)=\frac{1}{2}\int_0^1 (f'')^2$$
$$\int_0^1 f^2\leq \int_0^1 \big(x\cdot\int_0^1 (f')^2\big)=\frac{1}{2}\int_0^1 (f')^2\leq \frac{1}{4}\int_0^1 (f'')^2 $$
therefore $$\int_0^1 |(f^2(x))'| \,dx\leq \frac{1}{\sqrt{2}} \int_0^1 (f'')^2$$
A: By the Fundamental Theorem of Calculus
$$
f'(t)=\int_0^tf''(x)\,\mathrm{d}x\tag1
$$
and therefore,
$$
\begin{align}
f(t)
&=\int_0^tf'(x)\,\mathrm{d}x\tag{2a}\\
&=\int_0^t\int_0^xf''(y)\,\mathrm{d}y\,\mathrm{d}x\tag{2b}\\
&=\int_0^t\int_y^tf''(y)\,\mathrm{d}x\,\mathrm{d}y\tag{2c}\\
&=\int_0^t(t-y)f''(y)\,\mathrm{d}y\tag{2d}
\end{align}
$$
Explanation:
$\text{(2a)}$: Fundamental Theorem of Calculus
$\text{(2b)}$: apply $(1)$
$\text{(2c)}$: change order of integration
$\text{(2d)}$: evaluate inner integral
Thus,
$$
\begin{align}
\int_0^1\left|\left(f(x)^2\right)'\right|\,\mathrm{d}x
&=2\int_0^1\left|f(t)f'(t)\right|\,\mathrm{d}t\tag{3a}\\
&=2\int_0^1\left|\int_0^t(t-x)f''(x)\,\mathrm{d}x\int_0^tf''(x)\,\mathrm{d}x\right|\,\mathrm{d}t\tag{3b}\\
&\le2\int_0^1\|t-x\|_{L^2[0,t]}\|1\|_{L^2[0,t]}\|f''\|_2^2\,\mathrm{d}t\tag{3c}\\
&=2\int_0^1\frac{t^2}{\sqrt3}\,\mathrm{d}t\,\|f''\|_2^2\tag{3d}\\[3pt]
&=\frac2{3\sqrt3}\|f''\|_2^2\tag{3e}
\end{align}
$$
Explanation:
$\text{(3a)}$: Chain Rule
$\text{(3b)}$: apply $(1)$ and $(2)$
$\text{(3c)}$: Cauchy-Schwarz
$\text{(3d)}$: evaluate the $L^2[0,t]$ norms:
$\phantom{\text{(3d):}}$ $\|t-x\|_{L^2[0,t]}=\frac{t^{3/2}}{\sqrt3}$ and $\|1\|_{L^2[0,t]}=t^{1/2}$
$\text{(3e)}$: evaluate the integral
