Let $0\leq x < 1$, $1 \leq p < \infty$ and $q$ be the conjugate exponent defined by $$1/p + 1/q = 1.$$

I am looking for a nice proof that

$$ \frac{(1-x^p)^{1/p}(1-x^q)^{1/q}}{(1-x)(1+x^c)^{1/c}} \geq 1,$$ where $c$ is defined via $$2^{1/c} = p^{1/p} q^{1/q}.$$

The number $c$ is defined so that the inequality holds as we take a limit as $x \to 1$. I checked it for various values of $p$ graphically and it seems to be true. It holds with equality in the case $p=q=2$.

Edit: Here is a graph, for $p=20$.


The blue line is the graph of $$f(x) = \frac{(1-x^p)^{1/p} (1-x^q)^{1/q}}{1-x}.$$ The orange line is $(1+x^c)^{1/c}$, as in the original question. The green line is $2^{1/c - 1}(1+x)$ which was proved to be a lower bound for $f$ by Willie Wong in the comments.

  • 2
    $\begingroup$ It may be worth noting that if we replace the denominator by $2^{1/c - 1} (1-x)(1+x)$ (which is $\leq (1-x)(1 + x^c)^{1/c}$ so doesn't imply the desired the inequality), the inequality would be immediately be implied by Holder's inequality on the integral $\int_x^1 s^2~ ds/s$. So you are looking for something sharper that captures a bit of the error of naive applications of Holder. $\endgroup$ – Willie Wong Oct 7 '19 at 15:41
  • 3
    $\begingroup$ It seems interesting enough for MO. In the past, I got stuck in similar inequalities which were valuable to our research; e.g. see this post mathoverflow.net/questions/246919/… $\endgroup$ – T. Amdeberhan Oct 7 '19 at 18:58
  • 2
    $\begingroup$ @user64494, the "curious inequality" in the link is directly related to a problem arising in potential theory about whether animals should gather close to each other in order to decrease the total rate of heat loss. See mathoverflow.net/questions/217530/… $\endgroup$ – Paata Ivanishvili Oct 7 '19 at 21:00
  • 7
    $\begingroup$ What exactly is wrong with art for art's sake? $\endgroup$ – Lucia Oct 8 '19 at 12:13
  • 3
    $\begingroup$ @user64494 Then why spend time on this site decrying it? If other mathematicians find it interesting or useful, who are you to declare that it is not appropriate for this site? $\endgroup$ – Yemon Choi Oct 11 '19 at 2:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.