Let $f\colon[-1,1]\to\mathbb{R}$ be a continuous function. Prove that $$ 2\int_{-1}^{1}f(x)^2\: dx - \left(\int_{-1}^{1}f(x)\: dx\right)^2 \ge 3\,\left(\,\int\limits_{-1}^{1}x\,f(x)\: dx\right)^2 $$
I found something similar:
Integral Inequality $\int\limits_0^1f^2(x)dx\geq12\left( \int\limits_0^1xf(x)dx\right)^2.$
but don't know if estimation of that sort can be applied here.
On
It is useful to exploit the fact that Legendre polynomials give a complete orthogonal base of $L^2(-1,1)$ with the usual inner product. Given: $$ f(x) = \sum_{n\geq 0} a_n P_n(x) \tag{1}$$ we have: $$ \begin{array}{llcl}\text{(Parseval's identity)}&\displaystyle\int_{-1}^{1}f(x)^2\,dx &=& \displaystyle2\sum_{n\geq 0}\frac{a_n^2}{2n+1},\\ (P_0(x)=1)&\displaystyle\int_{-1}^{1} f(x)\,dx &=& \displaystyle2a_0,\tag{2}\\(P_1(x)=x)& \displaystyle\int_{-1}^{1} x\,f(x)\,dx &=& \displaystyle\frac{2}{3}a_1\end{array}$$
hence the original inequality is equivalent to: $$ 4\sum_{n\geq 0}\frac{a_n^2}{2n+1}-4a_0^2\geq \frac{4}{3}a_1^2 \tag{3} $$ or to:
$$ \sum_{n\geq 2}\frac{a_n^2}{2n+1}\geq 0 \tag{4}$$
that is trivial. We also have that equality is attained only by linear functions, $f(x)=a+bx$.
Let $\displaystyle\;\bar{f} = \frac12\int_{-1}^1 f(x) dx\;$ and $g(x) = f(x) - \bar{f}$. It is easy to check $$\int_{-1}^1 g(x) dx = 0 \quad\text{ and }\quad \int_{-1}^1 x g(x) dx = \int_{-1}^1 xf(x) dx $$ The LHS of inequality at hand can be rewritten as $$\verb/LHS/ = 2\int_{-1}^1 (g(x) + \bar{f})^2 dx - 4\bar{f}^2 = 2\int_{-1}^1 g(x)^2 dx = 3\left(\int_{-1}^1 x^2 dx\right)\left(\int_{-1}^1 g(x)^2 dx\right) $$ Apply Cauchy Schwarz to $x$ and $g(x)$ over $[-1,1]$, we find
$$\verb/LHS/ \ge 3 \left(\int_{-1}^1 xg(x)\right)^2 = 3 \left(\int_{-1}^1 xf(x) dx\right)^2 = \verb/RHS/ $$