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

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

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

• @Juho, if you managed to complete the question, consider posting your solution here for others to benefit. :-) – Zain Patel Jul 22 '15 at 22:33
• @Juho The AM-GM step gives an inequality that is not in the direction that you desire. – Kim Jong Un Jul 22 '15 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 :) – Hulkster Jul 23 '15 at 0:04
• This is tagged "contest-math". Could you please give the source contest from which the problem was taken? – Carl Mummert Nov 27 '16 at 23:18
• @CarlMummert As I stated, I found this problem by trial and error. Contest-math tag says "Problems from or inspired by mathematics competitions". My inspiration was this problem math.stackexchange.com/questions/707212 – Hulkster Nov 28 '16 at 15:58

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? – Hulkster Jul 23 '15 at 9:34
• See the edit above. – Batominovski Jul 23 '15 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/…). – Batominovski Jul 23 '15 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. – Hulkster Dec 7 '16 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. – Batominovski Dec 7 '16 at 11:58