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Is it true that the following inequality holds for any continuous, non-negative, and concave function $f:[0,1]\to \mathbb{R}$? $$\frac{3}{2}\,f(0)\int_0^1 xf(x)dx\leq \left(\int_0^1f(x)dx\right)^2.$$

Some thoughts. By homogeneity, we may assume that $f(0)=1$ (the case $f(0)=0$ is trivial).

Moreover, by concavity, for $x\in[0,1]$, $$f(x)=f((1-x)\cdot 0+x\cdot 1)\geq (1-x)f(0)+xf(1)\geq 1-x$$ because $f(1)\geq 0$.

Hence the function $h(x):=f(x)-(1-x)$ is continuous, non-negative and concave on $[0,1]$ and the inequality is equivalent to $$\frac{3}{2}\int_0^1 xh(x)dx\leq \left(\int_0^1h(x)dx\right)^2+\int_0^1h(x)dx,$$ that is $$\int_0^1 \left(\frac{3}{2}x-1\right)h(x)dx\leq \left(\int_0^1h(x)dx\right)^2.$$ So it suffices to show $$\int_0^1 \left(x-\frac{2}{3}\right)h(x)dx\leq 0$$ where $h(x)$ is a continuous, non-negative concave function on $[0,1]$ and $h(0)=0$.

Any hints or references are welcome.

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Finally, I managed to conclude the proof. We will show that $$\int_{2/3}^1 \left(x-\frac{2}{3}\right)h(x)dx\leq \int_0^{2/3} \left(\frac{2}{3}-x\right)h(x)dx.$$ Since $h$ is concave and $h(0)=0$ then $$m'x\leq h(x)\quad \mbox{for $x\in[0,2/3]$}$$ where $y=m'x$ is the secant line thru $(0,0)$ and $(2/3,h(2/3))$. Moreover $$h(x)\leq mx+q\quad \mbox{for $x\in[0,1]$}$$ where $y=mx+q$ is the tangent line at $2/3$. Then $$\int_{2/3}^1 \left(x-\frac{2}{3}\right)h(x)dx\leq \int_{2/3}^1 \left(x-\frac{2}{3}\right)(mx+q)=\frac{4m}{81}+\frac{q}{18}$$ and $$\int_0^{2/3} \left(\frac{2}{3}-x\right)h(x)dx\geq \int_0^{2/3} \left(\frac{2}{3}-x\right)(m'x)dx=\frac{4m'}{81}.$$ So it suffices to verify that $$\frac{4m'}{81}\geq \frac{4m}{81}+\frac{q}{18} \quad \Leftrightarrow\quad m'-m\geq \frac{9q}{8}$$ which holds because $q\geq h(0)=0$ and $m'(2/3)=m(2/3)+q$ implies $m'-m=3q/2$.

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