TITLE: integral inequality for $f(x)$ and $f(\sqrt{x})$
QUESTION [9 upvotes]: Show that if $f(x)\in [0;1]$, $f\in C$ and $\int\limits_{1}^{+\infty}f(t)dt=A$ then $\int\limits_{1}^{+\infty}tf(t)dt>\frac{A^2}{2}$
I only have noticed two small things:

If $A=1$ inequality is obvious, as expectation of random variable, that takes values greater than $1$
$\int\limits_{1}^{+\infty}tf(t)dt = \dfrac{1}{2}\int\limits_{1}^{+\infty}f(\sqrt{t})dt$

Please, provide hint for next steps
Also, some time left I obtain:
$$
A^2 = \left[\int\limits_{1}^{+\infty}f(x)x\dfrac{1}{x}dx\right]^2 \leq
\int\limits_{1}^{+\infty}\left(f(x)x\right)^2dx\int\limits_{1}^{+\infty}\left(\frac{1}{x}\right)^2dx\leq \int\limits_{1}^{+\infty}f(x)x^2dx
$$

REPLY [6 votes]: $$
\begin{aligned}
\int\limits_{1+A}^\infty xf(x)\,dx &\geq (1 + A)\int\limits_{1+A}^\infty f(x)\,dx\\
&= (1 + A)\Bigl[A - \int\limits_{1}^{1+A}f(x)\,dx\Bigr]\\
&= (1+A)\int\limits_{1}^{1+A}\underbrace{(1-f(x))}_{\geq 0}\,dx \\
&\geq \int\limits_{1}^{1+A}x(1-f(x))\,dx
\end{aligned}
$$
So
$$
\int\limits_{1}^\infty xf(x)\,dx \geq \int\limits_{1}^{1+A}x\,dx.
$$