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. 
 A: Apply The Cauchy-Schwarz Inequality for Integrals on the right side.
A: If $f$ is even, then $xf(x)$ is odd so its integral over $[-1,1]$ is zero and the inequality boils down to 
$$
\int_{-1}^1 f^2(x)dx \geq \frac{1}{2}\left( \int_{-1}^1 f(x)dx\right)^2
$$
and the result follows from Cauchy-Schwarz.
If $f$ is odd, then its integral over $[-1,1]$ is zero and the inequality becomes
$$
\int_{-1}^1 f^2(x)dx \geq \frac{3}{2}\left( \int_{-1}^1 xf(x)dx\right)^2
$$
and the result follows again from Cauchy-Schwarz.
Now write $f=g+h$ with $g$ even and $h$ odd.
We find
$$
\int_{-1}^1 (g+h)^2=\int g^2+\int h^2 + 2\int gh= \int g^2+\int h^2 
$$
since $gh$ is odd.
Also
$$
\left( \int g+h\right)^2=\left( \int g\right)^2+ \left( \int h\right)^2+ 2 \int g\int h=\left( \int g\right)^2
$$
since $h$ is odd and $\int h=0$.
Finally, we have
$$
\left( \int x(g+h)\right)^2= \left( \int xg\right)^2 + \left( \int xh\right)^2+ 2\int xg\int xh=\left( \int xh\right)^2
$$
since $xg(x)$ is odd and $\int xg=0$.
It only remains to add the two inequalities for $g$ and $h$, to obtain the one for $f$.
A: By Cauchy-Schwarz we have
$$\left(\int_{-1}^1f(x)dx\right)^2 \leq 2 \left(\int_{-1}^1f^2(x)dx\right)$$
$$\left(\int_{-1}^1xf(x)dx\right)^2 \leq \left(\int_{-1}^1x^2dx\right) \left(\int_{-1}^1f^2(x)dx\right)=\frac{2}{3} \left(\int_{-1}^1f^2(x)dx\right)$$
This proves the inequalities for odd and even functions.
Then use  SebastienB's hint.  
