341This is quite straighforward isnt it? This technique is carried over from algebraic inequalities:
\int_0^1 f^2(x) (1+x^2) \ dx \int_0^1 \frac{1}{1+x^2} \ dx \ge \left( \int_0^1 f(x) \ dx \right)^2
implies that the minimum is \int_0^1 f^2(x) (1+x^2) \ dx \ge \frac{4}{\pi}
From the equality condition for the Schwarz Bunyakovsky and the given condition , we have f(x) = \frac{4}{\pi} \frac{1}{1+x^2} [the other choice is the negative of this function, which is obvously not admissible]
