Minimum of integral Let $S$ be the set of all integrable on $[0,1]$ such that $$\int\limits_0^1f(x)dx=\int\limits_0^1xf(x)dx+1=3.$$
Prove that $S$ is infinite and evaluate $$\min\limits_{f\in S}\int\limits_0^1f^2(x)dx.$$

We have $3 = \int_0^1 f(x) \cdot \left(x +\frac{1}{3} \right) \, dx$. Let's use Cauchy–Schwarz inequality:
$$3 = \int_0^1 f(x)  \cdot \left(x +\frac{1}{3} \right) \, dx \le \sqrt{ \int_0^1 f^2(x) \, dx} \sqrt{\int_0^1 \left(x +\frac{1}{3} \right)^2 dx} = \sqrt{\frac{7}{9}} \sqrt{ \int_0^1 f^2(x) \, dx}$$
Hence:
$$\int_0^1 f^2(x) \, dx \ge \left( \frac{9 \cdot 3}{\sqrt{7}} \right)^2 = \frac{81}{7} \approx 11.57$$
Edit
$$2 = \int_0^1 x f(x) \, dx \le \sqrt{ \int_0^1 f^2(x) \, dx} \sqrt{\int_0^1 x^2 dx} = \frac{1}{\sqrt{3}} \sqrt{ \int_0^1 f^2(x) \, dx}$$
So:
$$\int_0^1 f^2(x) \, dx \ge 12 $$
And equality holds if $f(x) = 6x$
