I found the following question on another forum and unfortunately I don't have any additional information about it:
Let $f$ be non-negative and square integrable on $[0,1]$. Prove that
$$\left(\int_0^1 xf(x)\text{d}x\right)\left(\int_0^1 f^2 (x)\text{d}x\right)\geq\frac{4}{9}\left(\int_0^1 f(x)\text{d}x\right)^3.$$
I tried the usual Cauchy inequality tricks but nothing worked. I don't know if the inequality is tight, but I couldn't find an equality case.
Best Answer
Remarks. This is a variant of something I wrote some years ago. I learned this trick from @fedja (in other words, the solution is due to @fedja).
Proof.
We have, for any $b > 0, x \ge 0$, $$b^2 x + \frac{1}{b}(b - b^3 x)^{+} \ge 1 \tag{1}$$ where $(y)^{+} = \max(0, y)$.
(Note: If $b^2 x \ge 1$, (1) is true. If $b^2x < 1$, then $b^2 x + \frac{1}{b}(b - b^3 x)^{+} = b^2x + \frac{1}{b}(b - b^3x) = 1$.)
Using (1) and Cauchy-Bunyakovsky-Schwarz inequality, we have, for any $b > 0$, \begin{align*} \int_0^1 f(x) \,\mathrm{d} x &\le \int_0^1 \left(b^2 x + \frac{1}{b}(b - b^3 x)^{+}\right)f(x)\,\mathrm{d} x\\ &= b^2 \int_0^1 xf(x)\,\mathrm{d} x + \frac{1}{b}\int_0^1 (b - b^3 x)^{+} f(x)\,\mathrm{d} x\\ &\le b^2 \int_0^1 xf(x)\,\mathrm{d} x + \frac{1}{b}\sqrt{\int_0^1 \left[(b - b^3 x)^{+}\right]^2 \,\mathrm{d} x}\sqrt{\int_0^1 f^2(x)\,\mathrm{d} x}\\ &\le b^2 \int_0^1 xf(x)\,\mathrm{d} x + \frac{1}{b}\sqrt{\frac13}\sqrt{\int_0^1 f^2(x)\,\mathrm{d} x} \tag{2} \end{align*} where we use $$\int_0^1 \left[(b - b^3 x)^{+}\right]^2 \,\mathrm{d} x \le \int_0^{1/b^2} (b - b^3x)^2 \,\mathrm{d} x = \frac13. \tag{3} $$
Let $$A := \int_0^1 xf(x)\,\mathrm{d} x, \quad B := \sqrt{\frac13}\sqrt{\int_0^1 f^2(x)\,\mathrm{d} x}.$$ Since (2) holds for any $b > 0$, we choose $$b = \sqrt[3]{\frac12 \cdot \frac{1}{A} \cdot B}$$ to obtain $\int_0^1 f(x) \,\mathrm{d} x \le \frac32\sqrt[3]{2AB^2}$ or $$\int_0^1 f(x) \,\mathrm{d} x \le \frac32\sqrt[3]{\frac23}\sqrt[3]{\int_0^1 xf(x)\,\mathrm{d} x}\sqrt[3]{\int_0^1 f^2(x)\,\mathrm{d} x}$$ which results in $$\left(\int_0^1 xf(x)\,\mathrm{d}x\right)\left(\int_0^1 f^2 (x)\,\mathrm{d}x\right)\ge\frac{4}{9}\left(\int_0^1 f(x)\mathrm{d}x\right)^3.$$
We are done.