# Integral inequality

Let $f:[0,1]\longrightarrow \mathbb{R}$ be a continuous function such that $$\int_{0}^{1}f(x)dx=0.$$ Is the following inequality true? $$2\left(\int_{0}^{1}xf(x)dx\right)^2\leq \int_{0}^{1}(1-x^2)f^2(x)dx,$$ where $f^2(x)=f(x)\cdot f(x), \forall x \in \mathbb{R}$. Thank you very much!

From the Cauchy-Schwarz Inequality and from the condition $\int_0^1\,f(x)\,\text{d}x=0$, we get \begin{align} \left(\int_0^1\,x\,f(x)\,\text{d}x\right)^2 &=\left(\int_0^1\,\left(x-1\right)\,f(x)\,\text{d}x\right)^2 \\ &\leq{\left(\int_0^1\,\frac{\left(x-1\right)^2}{1-x^2}\,\text{d}x\right)\left(\int_0^1\,\left(1-x^2\right)\,\big(f(x)\big)^2\,\text{d}x\right)} \\ &={\big(2\ln(2)-1\big)\left(\int_0^1\,\left(1-x^2\right)\,\big(f(x)\big)^2\,\text{d}x\right)} \\ &\leq \frac{1}{2}\left(\int_0^1\,\left(1-x^2\right)\,\big(f(x)\big)^2\,\text{d}x\right)\,. \end{align} The equality holds iff $f(x)=0$ for almost all $x\in[0,1]$.