For $a_1,...,a_n,b_1,...,b_n>0,\quad$ define $a:=\sum a_i,\ b:=\sum b_i,\ s:=\sum \sqrt{a_ib_i}$.

Is the following inequality true?: $${\frac{\Bigl(\prod a_i^{a_i}\Bigr)^\frac1a}a \cdot \frac{\left(\prod b_i^{b_i}\right)^\frac1b}b \le\left(\frac{\prod (a_ib_i)^{ \sqrt{a_ib_i}}}{s^{2s}}\right)^\frac s{ab}}.$$ If true, how can I prove it? And if not true, is there a counter-example?

An alternate formulation of this inequality, as suggested by cardinal is

$$\sum_i x_i^2 \log x_i^2 + \sum_i y_i^2 \log y_i^2 \leq \rho^2 \sum_i \frac{x_iy_i}{\rho} \log \frac{x_iy_i}{\rho} \; ,$$

in which $x_i,y_i>0$, $\sum_i x_i^2 = \sum_i y_i^2 = 1$ and $\sum_i x_i y_i = \rho$. It can be found by taking the logarithm of the original expression and defining $x_i=\sqrt{a_i/a}$ and $y_i=\sqrt{b_i/b}$.

  • $\begingroup$ My answer is wrong. I should have done the limit by hand... $\endgroup$ May 26, 2011 at 22:00
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    $\begingroup$ Let $u_i = a_i / a$, $v_i = b_i / b$ and $w_i = \sqrt{a_i b_i} / s$. Then, the inequality in the question is equivalent to the inequality $\sum u_i \log u_i + \sum v_i \log v_i \leq \frac{2 s^2}{a b} \sum w_i \log w_i$. Note that $(u_i)$, $(v_i)$ and $(w_i)$ are all in the $(n-1)$-simplex and so correspond to discrete probability measures on $\{1,\ldots,n\}$. Probabilistically, we might rewrite this as $H(u) + H(v) \geq \frac{2 s^2}{a b} H(w)$ where $H(\cdot)$ denotes the Shannon entropy. $\endgroup$
    – cardinal
    May 27, 2011 at 13:11
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    $\begingroup$ Note also that there is a, perhaps, interesting element of symmetry at play here. In both the OP's question and my previous remark, the left-hand side is permutation invariant with respect to the relative ordering of the coordinates of $(a_i)$ and $(b_i)$, whereas the right-hand side is not. In particular, let $s_\pi = \sum \sqrt{a_i b_{\pi(i)}}$, $w_{i,\pi} = \sqrt{a_i b_{\pi(i)})}/s_\pi$, and $w_\pi = (w_{i,\pi})$. Then $H(u) + H(v) \geq \frac{2}{n! ab} \sum_{\pi \in S_n} s_\pi^2 H(w_\pi)$ and $H(u) + H(v) \geq \frac{2}{ab} \max_{\pi \in S_n} s_\pi^2 H(w_\pi)$. $\endgroup$
    – cardinal
    May 28, 2011 at 6:31
  • $\begingroup$ Yes, exactly... May we prove it for some small values of $n$? e.g. $n=3$ or $n=4$. $\endgroup$ May 30, 2011 at 5:26
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    $\begingroup$ @Amir Where did this problem come from? $\endgroup$
    – JavaMan
    May 30, 2011 at 21:11

3 Answers 3


This is slightly too long to be comment, though this is not an answer:

Edit: I am providing a few revisions, but I still have not found a proof.

My previous comment was not quite correct. I "rearranged" the inequality by raising both sides of the inequality to certain powers, and I did not take into account that we do not know a priori whether any of the terms are larger or smaller than $1$. Depending on the size of each term (in particular, whether it is smaller or larger than $1$), exponentiating both sides could potentially reverse the inequality. I have returned the question to its original form as suggested by the OP, and I have added one more observation.

Let me first say that I have been working on this question for a bit, and though I have not yet resolved it, I have been having fun trying!

Now, to emphasize the dependence on $n$, let's set

$$ \alpha_n = \sum_{i=1}^n a_i \qquad \beta_n = \sum_{i=1}^n b_i, \qquad \sigma_n = \sum_{i=1}^n c_i, $$

where $c_i = \sqrt{a_i b_i}$. Further, let's put

$$ A_n = \prod_{i=1}^n (a_i)^{a_i}, \qquad B_n = \prod_{i=1}^n (b_i)^{b_i}, \qquad S_n = \prod_{i=1}^n (c_i)^{c_i}. $$

Our goal is to show:

\begin{equation} (1) \hspace{1in} \left(\frac{A_n}{(\alpha_n)^{\alpha_n}}\right)^{\frac{1}{\alpha_n}} \cdot \left(\frac{B_n}{(\beta_n)^{\beta_n}} \right)^{\frac{1}{\beta_n}} \leq \left(\frac{S_n}{(\sigma_n)^{\sigma_n}}\right)^{\frac{2\sigma_n}{\alpha_n \beta_n}} \end{equation}

A few pedestrian observations:

  • If $a_i = b_i$ for $i = 1, \dots , n$ (which forces $c_i = a_i = b_i$), then $A_n = B_n = S_n$, we also have $\alpha_n = \beta_n = \sigma_n$, and (1) holds in this case.

  • Note that $2c_i \leq a_i + b_i$ as $2xy \leq x^2 + y^2$ for all real numbers $x, y$. Hence, $2\sigma_n \leq \alpha_n + \beta_n$. Furthermore, Cauchy-Schwarz gives $\sigma_n^2 \leq \alpha_n \beta_n$. Both of these observations imply that $(\sigma_n + 1)^2 \leq (\alpha_n + 1)(\beta_n + 1)$.

I would imagine that with enough creativity, one may find a proof of the inequality involving convexity or a simple application of the AM-GM inequality (which I suppose is much the same thing!).

I have been unable to prove the inequality even in the case $n = 2$, when I assume $\alpha_n = \beta_n = 1$. I am not hopeful for a proof of the general case.

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    $\begingroup$ Thanks for the start! I'm still waiting for a solution to this problem... $\endgroup$ May 28, 2011 at 5:13
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    $\begingroup$ @Amir: And I'm still thinking about it (: $\endgroup$
    – JavaMan
    May 28, 2011 at 15:10
  • $\begingroup$ BUMP this! Anyone? Just three days remaining... $\endgroup$ Jun 2, 2011 at 8:54
  • $\begingroup$ The bounty was going to be expired, so I gave it to DJC's post. Problem hasn't been solved yet, if anyone got anything, please share it with us. $\endgroup$ Jun 4, 2011 at 2:46
  • $\begingroup$ @Amir: I did not contribute to a solution of the problem at hand despite my efforts, and while I'm flattered to get the points, I simply don't deserve them. If there is a way to undo the points, I'm fine if they get taken away. $\endgroup$
    – JavaMan
    Jun 4, 2011 at 3:24

The inequality is true.

I posted cardinal's reformulation on Math Overflow, as I was very interested in seeing the problem resolved. A complete solution was provided by fedja, and it can be found here. Using the inequality $$ \sqrt{\frac{a_i b_i}{ab}}\leq \sum_{i=1}^n\sqrt{\frac{a_i b_i}{ab}}\leq \sqrt{\frac{a_i b_i}{ab}}+\sqrt{\frac{(a-a_i)(b- b_i)}{ab}},$$ which follows from Cauchy-Schwarz, fedja was able to consider the inequality entry-wise. Doing so, the problem was reduced to showing

(key inequality) For $x,y,z>0$ and $xy=z^2$, we have $$(1+x)\log(1+y)+(1+y)\log(1+x)\geq 2(1+z)\log(1+z).$$

Fedja gave a proof of this inequality in his answer, and the user Babylon5 gave an alternate proof on AoPS.


Not an answer :

We have the followig refinement for $a\geq 1$ and $x>0$ :

$$\left(x+a+2\right)\ln\left(x+a+2-\sqrt{\left(x+1\right)\left(a+1\right)}\right)-\ln\left(\left(x+1\right)\left(a+1\right)\right)+xa\left(\ln\left(\left(x+1\right)\left(a+1\right)\right)-2\left(1+\sqrt{\left(ax\right)^{-1}}\right)\ln\left(1+\sqrt{\left(ax\right)^{-1}}\right)\right)-\left(1+x\right)a\ln\left(a\right)-\left(1+a\right)x\ln\left(x\right)\geq 0$$

Obviously it seems harder .


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