Prove that $\frac{\int_0^1xf^2(x) \mathrm{d}x}{\int_0^1 xf(x) \mathrm{d}x}\le\frac{\int_0^1 f^2(x) \mathrm{d}x}{\int_0^1 f(x) \mathrm{d}x}$ [duplicate]

Let $f:[0,1]\rightarrow\mathbb{R_+}$ be a monotone decreasing function. We want to prove that

$$\frac{\int_0^1x(f(x))^2 \,\mathrm{d}x}{\int_0^1 xf(x) \,\mathrm{d}x}\le\frac{\int_0^1 (f(x))^2 \,\mathrm{d}x}{\int_0^1 f(x) \,\mathrm{d}x}$$

What would you propose here?

Thanks!

Sis.

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marked as duplicate by Norbert, Byron Schmuland, Amzoti, Dennis Gulko, Hagen von EitzenMar 1 '13 at 19:49

I also think of a possible generalization of the problem. For instance we could also ponder over this situation $$\frac{\int_0^1x^2(f(x))^2 \mathrm{d}x}{\int_0^1 x^2f(x) \mathrm{d}x}\le\frac{\int_0^1 x(f(x))^2 \mathrm{d}x}{\int_0^1 xf(x) \mathrm{d}x}$$ Is this true? No idea, but I'm very curious to find out. – user 1618033 Mar 1 '13 at 18:16
By "$f^2(x)$", do you mean $(f(x))^2$ or $f(f(x))$? The former you could write as $f(x)^2$ and everyone would understand that you don't mean $f(x^2)$, but there are reasons to prefer not to use the notation that you used. – Michael Hardy Mar 1 '13 at 18:19
@MichaelHardy: agree. I've changed the notation. – user 1618033 Mar 1 '13 at 18:21
@Norbert Good catch! I thought this problem looked familiar... – Byron Schmuland Mar 1 '13 at 18:40

Consider $F(t) = \int_0^t xf^2(x)dx\int_0^t f(x)dx - \int_0^t f^2(x)dx\int_0^t xf(x)dx$

Then $F(0) = 0$. We just prove $F(1)\le 0$.

We need to prove $F'(t)\le 0$,

while

$F'(t) = \int_0^t (t-x)f(x)f(t)(f(t)-f(x))dx$

Since $f$ is decreasing. it is done.

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+2: Nice approach. – copper.hat Mar 1 '13 at 18:35
@Yimin: I remember that you also came with a very nice proof for another question I posted some time ago. Glad to see you around :-) Thanks! (+1) – user 1618033 Mar 1 '13 at 18:41

Start by illustrating that it's true.

Rearrange it so that you are proving that

$$\frac{\int_0^1x(f(x))^2 \,\mathrm{d}x}{\int_0^1 (f(x))^2 \,\mathrm{d}x}\le\frac{\int_0^1 xf(x) \,\mathrm{d}x}{\int_0^1 f(x) \,\mathrm{d}x}$$

The left hand side is now the average of $x$ weighted by $f^2$, and the right hand side is the average of $x$ weighted by $f$. Since $f$ is decreasing, $f^2$ weights the average even further to low values of $x$ than does weighting via $f$. i.e LHS $\le$ RHS.

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does this argument also hold to a function $g(x) \geq 0$ instead of $x$? – clark Mar 1 '13 at 18:38
@clark No, I don't think so. E.g. if $g(x) = f(x) = 1-x$. – oks Mar 1 '13 at 18:59