Can you prove the following new inequality? I found it experimentally.

Prove that, for all $x_1,x_2,\ldots,x_n>0$, it holds that $$\sum_{i=1}^n\frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod\limits _{j=1}^nx_j}} \ge 1\,.$$

  • $\begingroup$ @Juho, if you managed to complete the question, consider posting your solution here for others to benefit. :-) $\endgroup$ – Zain Patel Jul 22 '15 at 22:33
  • $\begingroup$ @Juho The AM-GM step gives an inequality that is not in the direction that you desire. $\endgroup$ – Kim Jong Un Jul 22 '15 at 23:52
  • $\begingroup$ @Kim. That's good since I hate fast-food-solutions that are so common. If this question requires fresh ideas, it would be great :) $\endgroup$ – Hulkster Jul 23 '15 at 0:04
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    $\begingroup$ This is tagged "contest-math". Could you please give the source contest from which the problem was taken? $\endgroup$ – Carl Mummert Nov 27 '16 at 23:18
  • $\begingroup$ Hmmm... some users are trying to close this question, and I wonder why. Maybe this new inequality was too easy for them. $\endgroup$ – Hulkster Nov 29 '16 at 7:08

The inequality $\displaystyle\sum_{i=1}^n\frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod \limits_{j=1}^nx_j}} \ge 1$ is trivial given the claim below. Of course, the equality occurs if and only if $x_1=x_2=\ldots=x_n$.

Claim: For every $i=1,2,\ldots,n$, we have $\displaystyle\frac{x_i}{\sqrt[n]{x_i^n+\left(n^n-1\right)\,\prod\limits_{j=1}^n\,x_j}} \geq \frac{x_i^{1-\frac{1}{n^n}}}{\sum\limits_{j=1}^n\,x_j^{1-\frac{1}{n^n}}}$. The equality holds if and only if $x_1=x_2=\ldots=x_n$.

Proof: The required inequality is equivalent to $$\left(\sum_{j=1}^n\,x_j^{1-\frac{1}{n^n}}\right)^n-x_i^{n\left(1-\frac{1}{n^n}\right)}\geq \left(n^n-1\right)\,x_i^{-\frac{1}{n^{n-1}}}\,\prod_{j=1}^n\,x_j\,.$$ Note that the expansion of $\left(\sum\limits_{j=1}^n\,x_j^{1-\frac{1}{n^n}}\right)^n$ consists of $n^n$ terms of the form $x_{j_1}^{1-\frac{1}{n^n}}x_{j_2}^{1-\frac{1}{n^n}}\cdots x_{j_n}^{1-\frac{1}{n^n}}$, where $j_1,j_2,\ldots,j_n\in\{1,2,\ldots,n\}$. The product of these terms is equal to $$\left(\prod\limits_{j=1}^n\,x_j^{1-\frac{1}{n^n}}\right)^{n^n}\,.$$ If we take the term $x_i^{n\left(1-\frac{1}{n^n}\right)}$ out of the product, we get the product of $n^n-1$ terms from the expansion of $\left(\sum\limits_{j=1}^n\,x_j^{1-\frac{1}{n^n}}\right)^n-x_i^{n\left(1-\frac{1}{n^n}\right)}$, which is then equal to $$\frac{\left(\prod\limits_{j=1}^n\,x_j^{1-\frac{1}{n^n}}\right)^{n^n}}{x_i^{n\left(1-\frac{1}{n^n}\right)}}=x_i^{-\frac{1}{n^{n-1}}\left(n^n-1\right)}\,\prod\limits_{j=1}^n\,x_j^{n^n-1}\,.$$ By the AM-GM Inequality, $$\frac{\left(\sum\limits_{j=1}^n\,x_j^{1-\frac{1}{n^n}}\right)^n-x_i^{n\left(1-\frac{1}{n^n}\right)}}{n^n-1}\geq \left(x_i^{-\frac{1}{n^{n-1}}\left(n^n-1\right)}\,\prod\limits_{j=1}^n\,x_j^{n^n-1}\right)^{\frac{1}{n^n-1}}=x_i^{-\frac{1}{n^{n-1}}}\,\prod\limits_{j=1}^n\,x_j\,,$$ which is what we want. Hence, the claim is true. The equality case happens, due to the AM-GM Inequality, if and only if $x_1=x_2=\ldots=x_n$.

How did I get the exponent $1-\frac{1}{n^n}$?

I assumed it was $k$ at first, and the desired inequality was equivalent to $$\left(\sum\limits_{j=1}^n\,x_j^{k}\right)^n-x_i^{nk}\geq \left(n^n-1\right)\,x_i^{n(k-1)}\,\prod\limits_{j=1}^n\,x_j\,.$$ Then, the last inequality read $$\frac{\left(\sum\limits_{j=1}^n\,x_j^{k}\right)^n-x_i^{nk}}{n^n-1}\geq \left(x_i^{-nk}\,\prod\limits_{j=1}^n\,x_j^{n^nk}\right)^{\frac{1}{n^n-1}}\,,$$ the right-hand side of which I wanted to equal $x_i^{n(k-1)}\,\prod\limits_{j=1}^n\,x_j$. Therefore, $\frac{n^nk}{n^n-1}=1$ and $n(k-1)=-\frac{nk}{n^n-1}$, both of which gave me $k=1-\frac{1}{n^n}$.

  • $\begingroup$ OK, looks good! But how in Hell you invented those exponents $1-1/n^n$ in your claim? $\endgroup$ – Hulkster Jul 23 '15 at 9:34
  • $\begingroup$ See the edit above. $\endgroup$ – Batominovski Jul 23 '15 at 9:47
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    $\begingroup$ This is a technique called "isolated fudging." Another example is in IMO'2001#2 (artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/…). $\endgroup$ – Batominovski Jul 23 '15 at 10:07
  • $\begingroup$ I hastily accepted you answer before, but now I think I can follow. Can you imagine any other path with more traditional methods? Now your clever solution stands alone. $\endgroup$ – Hulkster Dec 7 '16 at 2:49
  • $\begingroup$ It is more natural to use Holder's Inequality, in a similar manner to a solution to IMO2001#2 which uses Holder's Inequality (i.e., Solution 1 in the given link above). I am very certain that it will work here too, but will not be trying because the algebra is much more involved. Maybe you can try that on your own. $\endgroup$ – Batominovski Dec 7 '16 at 11:58

By Holder $$\left(\sum_{i=1}^n\frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod\limits_{j=1}^nx_j}}\right)^n\sum_{i=1}^nx_i\left(x_i^n+(n^n-1)\prod\limits_{j=1}^nx_j\right)\geq\left(\sum_{i=1}^nx_i\right)^{n+1}$$ and it's enough to prove that: $$\left(\sum_{i=1}^nx_i\right)^{n+1}\geq\sum_{i=1}^nx_i^{n+1}+(n^n-1)\prod\limits_{j=1}^nx_j\sum_{i=1}^nx_i,$$ which is true by Muirhead because the term $\prod\limits_{j=1}^nx_j\sum\limits_{i=1}^nx_i$ is the smallest after deleting of $\sum\limits_{i=1}^nx_i^{n+1}.$


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