integral inequality for $f(x)$ and $f(\sqrt{x})$

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

Let $g(t)=f(t+1)$. We have to prove that for any $g\in C^0(\mathbb{R}^+)$ such that $g$ is bounded between $0$ and $1$ and $\int_{0}^{+\infty}g(t)\,dt = A$, $$ \int_{0}^{+\infty}(t+1)\,g(t)\,dt >\frac{A^2}{2}$$ holds. Let $\xi$ be the only real number such that $\int_{0}^{\xi}g(t)\,dt = \frac{A}{2}$ (the median). We have: $$ \int_{\xi}^{+\infty}(t+1)\,g(t)\,dt \geq (\xi+1)\int_{\xi}^{+\infty}g(t)\,dt = (\xi+1)\frac{A}{2}\tag{1}$$ but since $g(t)$ is bounded between $0$ and $1$, $\xi\geq \frac{A}{2}$ and the $RHS$ of $(1)$ is greater than $\color{red}{\large\frac{A^2}{4}}$.

Now we just need to refine a bit this argument and maybe consider what happens over $(0,\xi)$:

$$\int_{0}^{\xi}t g(t)\,dt = \left. t\,G(t)\right|_{0}^{\xi}-\int_{0}^{\xi}G(t)\,dt \geq \frac{\xi(A-\xi)}{2}\tag{2}$$ since $g\in(0,1)$ ensures $G(t)\leq t$. However, that does not improve $(1)$ if $\xi\geq A$.
Anyway, now we may consider $g_1(t)=g(t+\xi)$, that is a continuous and bounded (between $0$ and $1$) function whose integral over $\mathbb{R}^+$ is $\frac{A}{2}$. Let $\xi_1$ be the only positive real number such that $\int_{0}^{\xi_1}g_1(t)\,dt = \frac{A}{4}$. We have $\xi_1\geq \frac{A}{4}$ and: $$ \int_{\xi}^{+\infty}t\,g(t)\,dt = \frac{\xi A}{2}+\int_{0}^{+\infty}t\,g_1(t)\,dt\geq \frac{\xi A}{2}+\frac{\xi_1 A}{4}\tag{3}$$ so, by iterating this argument, we get: $$ \int_{\xi}^{+\infty} t\,g(t)\,dt \color{red}{\geq} \left(\frac{A}{2}\right)^2+\left(\frac{A}{4}\right)^2+\left(\frac{A}{8}\right)^2+\ldots = \color{red}{\frac{A^2}{3}}.\tag{4}$$