Inequality using integration by parts. I have shown easily that for a function $f$ such that $f(a)=f(b)=0, \int_{a}^{b}f^2=1$, that:
$$\int_{a}^{b}xf(x)f'(x)=-1/2$$
But how can I now show that:
$$\int_{a}^{b}f'(x)^2 \int_{a}^{b}x^2f(x)^2 > 1/4$$
I'm sure a hint will suffice here. I'm not sure what the natural choice is here as far as integration by parts goes.
 A: As noted by FH93, Cauchy-Schwarz shows that
$$
\begin{align}
\int_a^bf'(x)^2\,\mathrm{d}x\int_a^bx^2f(x)^2\,\mathrm{d}x
&\ge\left(\int_a^bxf(x)f'(x)\,\mathrm{d}x\right)^2\\
&=\left(-\frac12\right)^2\\
&=\frac14\tag{1}
\end{align}
$$
Cauchy-Schwarz can be rewritten as
$$
\begin{align}
0
&\le\int_a^b\left(\lambda xf(x)-\mu f'(x)\right)^2\,\mathrm{d}x\\
&=\lambda^2\int_a^bx^2f(x)^2\,\mathrm{d}x+\mu^2\int_a^bf'(x)^2\,\mathrm{d}x
-2\lambda\mu\int_a^bxf(x)f'(x)\,\mathrm{d}x\\
&=2\int_a^bx^2f(x)^2\,\mathrm{d}x\int_a^bf'(x)^2\,\mathrm{d}x\\
&-2\left(\int_a^bx^2f(x)^2\,\mathrm{d}x\int_a^bf'(x)^2\,\mathrm{d}x\right)^{1/2}\int_a^bxf(x)f'(x)\,\mathrm{d}x\tag{2}
\end{align}
$$
where we set $\lambda=\left(\int_a^bf'(x)^2\,\mathrm{d}x\right)^{1/2}$ and $\mu=\left(\int_a^bx^2f(x)^2\,\mathrm{d}x\right)^{1/2}$.
Thus, the inequality is strict unless
$$
\lambda xf(x)-\mu f'(x)=0\tag{3}
$$
which means
$$
\frac{f'(x)}{f(x)}=-2kx\tag{4}
$$
and therefore,
$$
f(x)=ce^{-kx^2}\tag{5}
$$
However, for any $f$ given in $(5)$, $f(a)=0$ or $f(b)=0$ could only hold if $c=0$, but then $f$ would be identically $0$ and then we would not have
$$
\int_a^bxf(x)f'(x)\,\mathrm{d}x=-\frac12\tag{6}
$$
Therefore, the inequality is strict.
A: Cauchy-Schwarz inequality trivialises it.
$\left(\displaystyle\int_a^b fg\right)^2 \leq \left(\displaystyle\int_a^b f^2 \right)\left(\displaystyle\int_a^b g^2\right)$
