xirti wrote: Prove that: $\int{-1}^1f^2(x)dx\dots$ What is $-1^1$ doing here? Is it just $-1$? Or you tried to write $\int_{-1}^1$ but made a mistake in $\LaTeX$?

I thinks he mean \[ \int_{-1}^{1} f^2(x) \, dx \geq \frac{1}{2} \left(\int_{-1}^1 f(x) \. dx\right)^2 + \frac{3}{2} \left( \int_{-1}^1 x f(x) \, dx \right)^2 \]

Yes sorry for the mistake. I have edited it. Please help me with this problem

In the inner product space $L^2[-1,1]$, the elements $\sqrt{\frac12}$ and $\sqrt{\frac32}x$ both have unit length, and are orthogonal. Therefore, $\sqrt{\frac12}\int_{-1}^1 f(x)\,dx$ is the component of $f$ in 1, and $\sqrt{\frac32}\int_{-1}^1 xf(x)\,dx$ is the component of $f$ in $x$. So, this amounts to saying that the norm of $f$ is at least as big as the sum of the squares of some of its components. The even and odd functions are just a convenient way of splitting into two orthogonal subspaces with direct sum equal to everything.

it is really nice approach, Jmerry