Inequality of continuous functions with integral

Question

Let M be a positive real number.Let $f:[0,\infty)\to[0,M] $ be a continuous function satisfying $$ \int\limits_{0}^\infty (1+x)f(x)dx<\infty.$$ Prove the following inequlity. $$\left( \int\limits_{0}^\infty f(x)dx \right)^2\le 4M\int\limits_{0}^\infty xf(x)dx.$$

Answer 1

Your proposal in the comments is close. Try this: $$ \frac{d}{dt} \left( \int_0^t f(x) dx \right)^2 = 2 \left( \int_0^t f(x) dx \right) \frac{d}{dt} \int_0^t f(x) dx = 2 f(t) \int_0^t f(x) dx \\ \le 2 f(t) \int_0^t M dx = 2 M t f(t). $$ Now integrate from $t=0$ to $t=T$: $$ \left( \int_0^T f(x) dx \right)^2 \le \int_0^T 2 M tf(t) dt $$ and then send $T \to \infty$ to get $$ \left( \int_0^\infty f(x) dx \right)^2 \le \int_0^\infty 2 M xf(x) dx. $$