Let $f:[0,1]\to[1,3]$ be continuous. Prove $$1 \leq \int_0^1 f(x)\,\mathrm dx \int_0^1 \frac{1}{f(x)}\, \mathrm dx\leq \frac{4}{3}.$$

The left is just Cauchy's inequality with integral form, but what's the right?


Since $f(x) + 3/f(x) \leq 4$ for all $x$, $$ \int_0^1 f(x)dx + 3\int_0^1 \frac{1}{f(x)}dx \leq 4 $$ Apply AM-GM inequality on LHS to get the result.

The equality holds when the value of $f$ is $1$ for the half of $[0,1]$ and $3$ for the other half, though it is not continuous.

  • $\begingroup$ That's fine! Thanks. $\endgroup$ – mengdie1982 May 19 at 12:28

Thanks for @JWL's hint. I will write down an entire proof for the problem. Please correct me if I'm wrong.


As for the left part of the inequality we want to prove, denote $$g(x):=\int_0^x f(t)dt\int_0^x\frac{1}{f(t)}dt-x^2,$$then it is equivalent to $g(1)\geq 0$. Notice that $g(0)=0$, and $$\begin{aligned} g'(x)&=f(x)\int_0^x\frac{1}{f(t)}dt+\frac{1}{f(x)}\int_0^x f(t)d-2x\\&=\int_0^x \frac{f(x)}{f(t)}+\frac{f(t)}{f(x)} dt-2x\\&\geq\int_0^x 2dt-2x\\&=0,\end{aligned}$$which shows that $g(x)$ is increasing monotonically. Thus $g(1)\geq g(0)=0$, which is just we want.

Let's turn to tackle the right hand part . Notice that ,under the condition $f(x) \in [1,3]$$$ f(x)+\frac{3}{f(x)}\leq 4.$$Therefore$$\int_0^1 f(x)+\frac{3}{f(x)}dx \leq \int_0^1 4 dx=4,$$which is $$\int_0^1 f(x)dx+3\int_0^1\frac{1}{f(x)}dx \leq 4.$$ As per AM-GM inequality, we obtain $$\int_0^1 f(x)dx+3\int_0^1\frac{1}{f(x)}dx\geq 2 \left[\int_0^1 f(x)dx \cdot 3\int_0^1\frac{1}{f(x)}dx\right]^{1/2},$$which gives$$\int_0^1 f(x)dx \cdot \int_0^1\frac{1}{f(x)}dx\leq \frac{4}{3}.$$

  • $\begingroup$ Last line should say 4/3 instead of 3/4. Otherwise I think it is fine $\endgroup$ – JustAnotherStackUser May 19 at 13:56
  • $\begingroup$ @MaximilianJanisch Thanks! Corrected! $\endgroup$ – mengdie1982 May 19 at 16:07

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