Show that $\left(\int_0^1 f(x)\,dx\right)^2\leq2\int_0^1x\, f(x)^2\,dx $ Let $f$ be a nondecreasing, integrable, function defined on $[0, 1]$. Show that $$\left(\int_0^1 f(x)\,dx\right)^2\leq2\int_0^1x \,f(x)^2\,dx $$
 A: Let's make some use of Chebyshev Integral Inequality:
$$\left(\int_0^1 f(x)~dx\right)^2\leq\int_0^1(f(x))^2~dx \tag1$$
$$\int_0^1(f(x))^2~dx=2\left(\int_0^1x ~dx\ \right) \left(\int_0^1 ((f(x))^2 ~dx\right)\leq2 \int_0^1x (f(x))^2~dx \tag2 $$
From $(1)$ and $(2)$ we obtain what we need
$$\left(\int_0^1 f(x)~dx\right)^2\leq2\int_0^1x (f(x))^2~dx$$
Q.E.D.
A: Here's another solution, via double integration: using the fact that $f$ is nondecreasing, we have $$\left(\int_{0}^{1} f(x) \, dx \right)^2 = \int_{0}^{1} \int_{0}^{1} f(x) f(y) \, dx \, dy $$ $$= \int_{0\le x \le y \le 1} f(x)f(y) \, dx\, dy + \int_{0\le y \le x \le 1} f(x)f(y) \, dx\, dy $$ $$\le \int_{0\le x \le y \le 1} f(y)^2 \, dx\, dy+\int_{0\le y \le x \le 1} f(x)^2 \, dx\, dy $$ $$ =2\int_{0\le y \le x \le 1} f(y)^2 \, dx \, dy= 2 \int_{0}^{1} \int_{0}^{x} f(y)^2 \, dy \, dx $$ $$\le 2\int_{0}^{1}  \int_{0}^{x} f(x)^2 \, dy \, dx = 2\int_{0}^{1} xf(x)^2\, dx$$ as desired.
A: 
Fact: Let $a$ and $b$ denote two nondecreasing functions on $[0,1]$, then $\int\limits_0^1ab\geqslant\int\limits_0^1a\cdot\int\limits_0^1b$.

To solve your question, apply this to the functions $a:x\mapsto2x$ and $b=f^2$, then apply Cauchy-Schwarz inequality to get $\int\limits_0^1b\geqslant\left(\int\limits_0^1f\right)^2$.
To prove the Fact recalled above, consider the function $c:(x,y)\mapsto(a(x)-a(y))(b(x)-b(y))$ defined on $[0,1]^2$ and note that $\iint\limits_{[0,1]^2} c=2\int\limits\limits_0^1 ab-2\int\limits_0^1 a\cdot\int\limits_0^1 b$ and that $c\geqslant0$ on $[0,1]^2$.
