Integral inequality $\int_{0}^{1}h(x)^2dx\int_{0}^{1}x^2h(x)^2dx\ge c\left(\int_{0}^{1}h(x)dx\right)^4$ Does there exist a number $c>0$ such that for every measurable function $h:[0,1]\to\mathbb{R}$, $h(x)\ge 0~\forall x$, we have $$\int_{0}^{1}h(x)^2dx\int_{0}^{1}x^2h(x)^2dx\ge c\left(\int_{0}^{1}h(x)dx\right)^4~~~~?$$
 A: The following is Problem 7.1 in "The Cauchy Schwarz Master Class"
by J. Michael Steele: Show
$$
\int_{-\infty}^{\infty}\left|f\left(t\right)\right|\, dt\leq8^{1/2}\cdot\left(\int_{-\infty}^{\infty}\left|f\left(t\right)\right|^{2}\, dt\right)^{1/4}\cdot\left(\int_{-\infty}^{\infty}\left|t\cdot f\left(t\right)\right|^{2}\, dt\right)^{1/4}.
$$
It is easy to see that this inequality implies your claim.
The proof goes as follows: We partition the domain of integration
into $T=\left(-t,t\right)$ and $T^{c}$. By using the Cauchy Schwarz
inequality, we get
$$
\int_{T}\left|f\left(t\right)\right|\, dt=\int_{T}1\cdot\left|f\left(t\right)\right|\, dt\leq\sqrt{\int_{T}1^{2}\, dt}\cdot\sqrt{\int_{T}\left|f\left(t\right)\right|^{2}\, dt}\leq\sqrt{2t}\cdot\sqrt{\int_{-\infty}^{\infty}\left|f\left(t\right)\right|^{2}\, dt}
$$
and
\begin{eqnarray*}
\int_{T^{c}}\left|f\left(t\right)\right|\, dt & = & \int_{T^{c}}\frac{1}{\left|t\right|}\cdot\left|t\cdot f\left(t\right)\right|\, dt\\
 & \leq & \sqrt{\int_{T^{c}}\frac{1}{t^{2}}\, dt}\cdot\sqrt{\int_{T^{c}}\left|t\cdot f\left(t\right)\right|^{2}\, dt}\\
 & \leq & \sqrt{\frac{2}{t}}\cdot\sqrt{\int_{-\infty}^{\infty}\left|t\cdot f\left(t\right)\right|^{2}\, dt}.
\end{eqnarray*}
Hence, 
\begin{eqnarray*}
\int_{-\infty}^{\infty}\left|f\left(t\right)\right|\, dt & = & \int_{T}\left|f\left(t\right)\right|\, dt+\int_{T^{c}}\left|f\left(t\right)\right|\, dt\\
 & \leq & \sqrt{2t}\cdot A^{1/2}+\sqrt{\frac{2}{t}}\cdot B^{1/2}=:\varphi\left(t\right)
\end{eqnarray*}
with $A:=\int_{-\infty}^{\infty}\left|f\left(t\right)\right|^{2}\, dt$
and $B:=\int_{-\infty}^{\infty}\left|t\cdot f\left(t\right)\right|^{2}\, dt$.
By setting $t_{0}=B^{1/4}/A^{1/4}$ (this choice can be found by optimizing
$\varphi$ using calculus), we get
$$
\varphi\left(t_{0}\right)=\sqrt{2}\frac{A^{1/2}B^{1/4}}{A^{1/4}}+\sqrt{2}\frac{B^{1/2}A^{1/4}}{B^{1/4}}=2\sqrt{2}A^{1/4}B^{1/4}=8^{1/2}A^{1/4}B^{1/4},
$$
which completes the proof.
This technique of splitting the domain of integration by introducing a parameter and optimizing with respect to this parameter in the end is worth remembering.
