Find $$ \min_{f \in \mathcal{A}} \int_{0}^{1} (1+x^{2})(f(x))^{2} dx $$ where $$ \mathcal{A}=\left\{ f \in C[a,b] \ | \ \int_{0}^{1} f(x) dx = 1 \right\}.$$
I used the Hölder's inequality to try to solve the problem but I do not know how to reduce the term with $(f (x))^{4}$, i.e, $$\int_{0}^{1} (1+x^{2})(f(x))^{2} dx \leq \left( \int_{0}^{1} (1+x^{2})dx \right)^{\frac{1}{2}} \left( \int_{0}^{1} (f(x))^{4} dx \right)^{\frac{1}{2}} $$ I appreciate any help.
We have $$ 1=\int_0^1 f(x)dx = \int_0^1 \frac{1}{\sqrt{1+x^2}} (\sqrt{1+x^2} f(x))dx $$ $$ \le \left(\int_0^1 \frac{1}{1+x^2} dx \right)^{1/2} \left(\int_0^1 (1+x^2) (f(x))^2 dx\right)^{1/2} $$ $$ =\sqrt{\frac{\pi}{4}} \left(\int_0^1 (1+x^2) (f(x))^2 dx\right)^{1/2}. $$ Thus $$ \frac{4}{\pi} \le \int_0^1 (1+x^2) (f(x))^2 dx. $$ For $f_{min}(x):= \frac{4}{\pi} \frac{1}{1+x^2}$ we have $$ \int_0^1f_{min}(x) dx=1, \quad \int_0^1 (1+x^2) (f_{min}(x))^2 dx = \frac{4}{\pi}. $$ Thus the minimum is $4/\pi$.
