$\left[\int_0^\infty x^2f(x)\,\mathrm{d} x\right]^2\leq \frac{8}{9}\int_0^\infty xf(x)\,\mathrm{d} x\cdot \int_0^\infty x^3f(x)\,\mathrm dx$ 
Let $\displaystyle f(x)$ defined on $\displaystyle [0,+\infty)$, continuously differentiable and decreasing, $\displaystyle \int_0^\infty x^3f(x)\,\mathrm{d} x$ converges. Show that
\begin{aligned}
\left[\int_0^\infty\!\!x^2f(x)\,\mathrm{ d} x\right]^2\!\leq
\frac{8}{9}\!\int_0^\infty\!\!xf(x)\,\mathrm{ d} x\cdot \!\!\int_0^\infty\!\!x^3f(x)\,\mathrm{ d} x.\end{aligned}

If $8/9$ is neglected, then we can prove it just by Schwarz. Oh, what the decreasing implies this nice inequality?
 A: By the assumption, we must have $f(\infty) := \lim_{x\to\infty} f(x) = 0$. We also note that $g(x) := -f'(x) \geq 0$ for all $x \geq 0$. So, for any $p > 0$,
\begin{align*}
\int_{0}^{\infty} x^p f(x) \, \mathrm{d}x
&= \int_{0}^{\infty} x^p \left( \int_{x}^{\infty} g(t) \, \mathrm{d}t \right) \, \mathrm{d}x \\
&= \int_{0}^{\infty} \left( \int_{0}^{t} x^p \, \mathrm{d}x \right) g(t) \, \mathrm{d}t
= \frac{1}{p+1} \int_{0}^{\infty} t^{p+1} g(t) \, \mathrm{d}t.
\end{align*}
Here, the second step follows from the Fubini–Tonelli theorem. We also remark that this is essentially the same as what we would have obtained from integration-by-parts (IbP) technique, although justifying IbP in this scenario is not so trivial.
Returning to the original question, now OP's inequality is equivalent to
$$ \left( \int_{0}^{\infty} t^3 g(t) \, \mathrm{d}t \right)^2
\leq \left( \int_{0}^{\infty} t^2 g(t) \, \mathrm{d}t \right) \left( \int_{0}^{\infty} t^4 g(t) \, \mathrm{d}t \right), $$
which is now a direct consequence of the Cauchy–Schwarz inequality.
