Let $f(x)$ is a continuous function in $[1;3]$ such that $\max\limits_{x\in[1;3]} f(x)=2$ and $\min\limits_{x\in[1;3]} f(x)=\dfrac{1}{2}$. Find the maximum value of $\displaystyle P=\int\limits_1^3 f(x)dx\int\limits_1^3\dfrac{1}{f(x)}dx$.

I think that $P$ achives the maximum value iff $f(x)$ is a linear function , where $f(x)=\dfrac{11}{4}-\dfrac{3}{4}x$ or $f(x)=\dfrac{3}{4}x-\dfrac{1}{4}$. In that cases, $P=\dfrac{20\ln 2}{3}$.

If applying the Cauchy-Schwarz inequality, we get $\displaystyle P=\int\limits_1^3 f(x)dx\int\limits_1^3\dfrac{1}{f(x)}dx\color{red}\ge \left(\int\limits_1^3 \sqrt{f(x)}\cdot\sqrt{\dfrac{1}{f(x)}}dx\right)^2$, and get stuck.

So who can help me?


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