How prove $\frac{\int_{0}^{1}xf^2(x)dx}{\int_{0}^{1}xf(x)dx}\le\frac{\int_{0}^{1}f^2(x)dx}{\int_{0}^{1}f(x)dx}$ [duplicate]

let $f(x)$be positive and decreasing on $[0,1]$ show that: $$\dfrac{\displaystyle\int_{0}^{1}xf^2(x)dx}{\displaystyle\int_{0}^{1}xf(x)dx}\le\dfrac{\displaystyle\int_{0}^{1}f^2(x)dx}{\displaystyle\int_{0}^{1}f(x)dx}$$

following is my poof: $$\Longleftrightarrow \int_{0}^{1}f^2(x)dx\cdot\int_{0}^{1}yf(y)dy-\int_{0}^{1}yf^2(y)dy\cdot\int_{0}^{1}f(x)dx\ge 0$$ then the $LHS$ is $$I=\int_{0}^{1}\int_{0}^{1}f(x)f(y)y[f(x)-f(y)]dxdy$$

and $x\longrightarrow y,y\longrightarrow x$,we have $$I=\int_{0}^{1}\int_{0}^{1}f(x)f(y)x[f(y)-f(x)]dxdy$$ add this $$2I=\int_{0}^{1}\int_{0}^{1}f(x)f(y)(y-x)[f(x)-f(y)]dxdy$$ since $f$ is decreaing on $[0,1]$,then we $$(y-x)[f(x)-f(y)]\ge 0,0\le x,y\le 1$$

so $I\ge 0$.

my question: This inequality have other nice methods?(and How many ?) such of the mean of theorem .

Thank you.

other my idea:

let $$F(x)=\int_{0}^{x}xf^2(y)dy,g(x)=\int_{0}^{x}f^2(y)dy$$ use cauchy mean of theorem, there exist $\xi\in (0,1)$, such that $$\dfrac{\displaystyle\int_{0}^{1}xf^2(x)dx}{\displaystyle\int_{0}^{1}xf(x)dx}=\dfrac{F(1)-F(0)}{g(1)-g(0)}=\dfrac{F'(\xi)}{g(\xi)}=\xi$$

following I can't work.Thank you everyone .