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Let $f\geq 0$ and be decreasing on $[0,1]$, prove that$$2\int_0^1 xf(x) dx\leq \left(\int_0^1 \sqrt{f(x)} dx\right)^2.$$

Usually $p$-norm is larger when $p$ is larger, but in this case there is a "weight" $x$ that is larger when $f$ is smaller ($f$ is decreasing).

So far, I've tried C-S, MVT (weighted or not), remainder of Taylor, etc. but nothing works.

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Let us make the observation that \begin{align} x\sqrt{f(x)} \leq \int^x_0 \sqrt{f(s)}\ ds \end{align} for all $0\leq x \leq 1$, since $f$ is decreasing. Then we see that \begin{align} xf(x) \leq \sqrt{f(x)} \int^x_0 \sqrt{f(s)}\ ds \ \ \implies& \ \ \int^1_0 xf(x)\ dx \leq \int^1_0 \left(\sqrt{f(x)}\int^x_0 \sqrt{f(s)}\ ds\right)\ dx\\ \implies& \ \ \int^1_0 xf(x)\ dx \leq \frac{1}{2}\int^1_0 \frac{d}{dx}\left( \int^x_0 \sqrt{f(s)}\ ds\right)^2\ dx \\ \implies&\ \ \int^1_0 xf(x)\ dx \leq \frac{1}{2} \left(\int^1_0\sqrt{f(s)}\ ds \right)^2. \end{align}

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