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Given that $f$ is continuous, decreasing and always positive on $[0,1]$, show that

$$ \frac{\operatorname{\Large\int}_0^1 \ xf^2(x)dx}{\operatorname{\Large\int}_0^1 xf (x)dx} \leq \frac{\operatorname{\Large\int}_0^1 f^2(x)dx}{\operatorname{\Large\int}_0^1 f (x)dx}. $$

I am trying to prove this but I do not get any lead, any hint is appreciated

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  • $\begingroup$ With the / symbol what do you mean? May be a division? $\endgroup$
    – Gabrielek
    Jan 14 '21 at 10:32
  • $\begingroup$ yes it is division $\endgroup$
    – user345777
    Jan 14 '21 at 10:36
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    $\begingroup$ Please avoid no-clue questions. $\endgroup$ Jan 14 '21 at 10:53
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\begin{align*} \int_0^1\int_0^1 & (x-y)f^2(x)f(y)\mathrm{d}x\mathrm{d}y = \int_0^1\int_0^y(x-y)f^2(x)f(y)\mathrm{d}x\mathrm{d}y\ + \\ & \int_0^1\int_0^x(x-y)f^2(x)f(y)\mathrm{d}y\mathrm{d}x = \int_0^1\int_0^y(x-y)\big[f^2(x)f(y)-f(x)f^2(y)\big]\mathrm{d}x\mathrm{d}y <0, \end{align*}

since $f$ is decreasing.

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  • $\begingroup$ how does this prove the statement $\endgroup$
    – user345777
    Jan 14 '21 at 11:20
  • $\begingroup$ Make a common denominator, you get two integrals from both sides both in dx. Change the variable of integration of one integral in y and due to Fubini's theorem you get a double integral on $[0,1]^2$. Then apply what @edwinFranks has written above. $\endgroup$
    – Gabrielek
    Jan 14 '21 at 11:30

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