For all integrable $f:[-1,1]\mapsto \mathbb{R}$ prove that $\int_{-1}^1f^2(x)\ge\frac12(\int_{-1}^1f(x))^2+\frac32(\int_{-1}^1xf(x))^2$ For all integrable  $f:[-1.1]\mapsto \mathbb{R}$ peove that $$\int_{-1}^1f^2(x)dx\ge\frac12\left(\int_{-1}^1f(x)dx\right)^2+\frac32\left(\int_{-1}^1xf(x)dx\right)^2$$
Thanks in advance. 

Apply The Cauchy-Schwarz Inequality for Integrals on the right side.
