For $\alpha>0$, let $\mathcal{S}_{\alpha}$ be the set of functions $f:[-\alpha,\alpha]\mapsto\mathbb{R}$ with the property $\int_{-\alpha}^{\alpha}(f(x))^2\mathrm dx=1$.
What is the $\sup$ of the set $\left(\int_{-\alpha}^{\alpha}f(x)\mathrm dx\right)^2+\left(\int_{-\alpha}^{\alpha}xf(x)\mathrm dx\right)^2+\left(\int_{-\alpha}^{\alpha}x^2f(x)\mathrm dx\right)^2$ for $f\in\mathcal{S}_{\alpha}$.
I first prove that
$$ \left( \int_{-\alpha}^{\alpha} \sqrt{x^2+1}|f(x)|\ \text{d}x \right)^2 \geq \left( \int_{-\alpha}^{\alpha} f(x)\ \text{d}x\right)^2 + \left( \int_{-\alpha}^{\alpha} xf(x)\ \text{d}x \right)^2 $$
and
$$ \left( \int_{-\alpha}^{\alpha} \sqrt{x^2+x^2+1}|f(x)|\ \text{d}x \right)^2 \geq \left( \int_{-\alpha}^{\alpha} f(x)\ \text{d}x\right)^2 + \left( \int_{-\alpha}^{\alpha} xf(x)\ \text{d}x \right)^2+ \left( \int_{-\alpha}^{\alpha} x^2f(x)\ \text{d}x \right)^2 $$ I am wondering if I can use Cauchy-Schwart at this stage.