$\sum\limits_{i=1}^n \frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod \limits_{j=1}^nx_j}} \ge 1$, for all $x_i>0.$

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 $$\sum_{i=1}^n\frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod\limits _{j=1}^nx_j}} \ge 1\,.$$

• @Juho, if you managed to complete the question, consider posting your solution here for others to benefit. :-) Jul 22, 2015 at 22:33
• @Juho The AM-GM step gives an inequality that is not in the direction that you desire. Jul 22, 2015 at 23:52
• @Kim. That's good since I hate fast-food-solutions that are so common. If this question requires fresh ideas, it would be great :) Jul 23, 2015 at 0:04
• This is tagged "contest-math". Could you please give the source contest from which the problem was taken? Nov 27, 2016 at 23:18
• Hmmm... some users are trying to close this question, and I wonder why. Maybe this new inequality was too easy for them. Nov 29, 2016 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 $k=1-\frac{1}{n^n}$.

• OK, looks good! But how in Hell you invented those exponents $1-1/n^n$ in your claim? Jul 23, 2015 at 9:34
• See the edit above. Jul 23, 2015 at 9:47
• This is a technique called "isolated fudging." Another example is in IMO'2001#2 (artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/…). Jul 23, 2015 at 10:07
• 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. Dec 7, 2016 at 2:49
• 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. Dec 7, 2016 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 $$\sum\limits_{i=1}^nx_i^{n+1}.$$