Minimum of integral through continuous functions $A$ is the set of all $f$ functions for which $f$ is continuous and $\int_0^1 f(x)dx=1$.
What is
$$min_{f\in A} \int_0^1(1+x^2)f^2(x)dx$$
and the corresponding $f$?
I tried integrating by parts with $du=f(x)$ and $v=(1+x^2)f(x)$ but I got nothing.
 A: Let $h(x)=x+\frac{x^3}{3}$ and $g(x)$ the inverse function of $h(x)$. Let $u=f\circ g$. We have:
$$ 1 = \int_{0}^{1} f(x)\,dx = \int_{0}^{4/3} f(g(t)) g'(t)\,dt = \int_{0}^{4/3} u(t)\frac{1}{1+g^2(t)}\,dt $$
and we want to find ($E$ is the set of functions fulfilling the previous constraint):
$$ \inf_{f\in L^2\cap E}\int_{0}^{1}(1+x^2)\,f(x)^2\,dx = \inf_{u\in L^2\cap E}\int_{0}^{4/3} u(t)^2\,dt.$$
If we consider a complete orthogonal base of $L^2\left(0,\frac{4}{3}\right)$ that contains $\frac{1}{1+g^2(t)}$, Parseval's inequality gives that the last infimum is indeed a minimum and it occurs at:
$$ u(t) = \frac{C}{1+g^2(t)},$$
but in such a case 
$$ f(x) = u(h(x)) = \frac{C}{1+g^2(h(x))} = \frac{C}{1+x^2} $$
and $C=\frac{4}{\pi}$ in order to meet $\int_{0}^{1}f(x)\,dx = 1$. That gives:
$$ \forall f\in L^2(0,1):\int_{0}^{1}f(x)\,dx=1,\qquad \int_{0}^{1}(1+x^2)\,f(x)^2\,dx \geq \int_{0}^{1}(1+x^2)\left(\frac{4}{\pi}\cdot\frac{1}{1+x^2}\right)^2\,dx=\color{red}{\frac{4}{\pi}}.$$
Obviously $C^0(0,1)\subset L^2(0,1)$ and $f(x)=\color{red}{\frac{4}{\pi}\cdot\frac{1}{1+x^2}}\in C^0(0,1)$, so the original problem is solved.

An alternative approach is through Lagrange multipliers. Since we are dealing with continuous functions, we may replace our integrals with Riemann sums and let $a_k = f\left(\frac{k}{n}\right) $. The infimum of
$$F(a_1,\ldots,a_n)=\frac{1}{n}\sum_{k=1}^{n}a_k^2\left(1+\frac{k^2}{n^2}\right) $$
over the set $a_1+a_2+\ldots+a_n=n$ occurs when
$$ \forall k\in[1,n],\qquad \lambda = \frac{\partial F}{\partial a_k} = 2a_k\left(1+\frac{k^2}{n^2}\right), $$
hence for
$$ a_k = \frac{C}{1+\left(\frac{k}{n}\right)^2} $$
that by getting rid of the discretization is equivalent to $f(x)=\frac{C}{1+x^2}$ just as before.

Now the easiest approach. By the Cauchy-Schwarz inequality:
$$ \left(\int_{0}^{1}f(x)\,dx\right)^2\leq \int_{0}^{1}\frac{dx}{1+x^2}\int_{0}^{1}(1+x^2)\,f(x)^2\,dx $$
hence $$\int_{0}^{1}(1+x^2)\,f(x)^2\,dx\geq\frac{4}{\pi}.$$ 
