Show that $(\int_0^1 f(x)~dx)^2\leq2\int_0^1x (f(x))^2~dx$

Let $f$ be a non-decreasing, integrable function defined on $[0, 1]$. Show that $$(\int_0^1 f(x)~dx)^2\leq2\int_0^1x (f(x))^2~dx$$

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for a reference for the inequality below en.wikipedia.org/wiki/Chebyshev's_sum_inequality – clark Aug 15 '12 at 12:43
@clark: I didn't notice your reference before finishing my post. :-) – user 1618033 Aug 15 '12 at 13:03
Nice question (+1) – user 1618033 Aug 15 '12 at 13:31
@Chris'sister: Thank you – A.D Aug 17 '12 at 20:24
Thank you all for help me so nicely. – A.D Aug 17 '12 at 20:25

2 Answers

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 $[0,1]^2$.

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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.

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(+1)nice procedure – Argha Aug 15 '12 at 14:34
@Ranabir: Thank you. I'm glad to know that my answers help. – user 1618033 Aug 15 '12 at 14:40