# inequality of integrals [closed]

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

• With the / symbol what do you mean? May be a division? Jan 14 '21 at 10:32
• yes it is division Jan 14 '21 at 10:36
• Please avoid no-clue questions. Jan 14 '21 at 10:53

\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.
• 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. Jan 14 '21 at 11:30