By Cauchy-Schwarz Inequality, we have \[\sum\limits_{i = 1}^n {\frac{{x_i^5}}{{{x_1} + {x_2} + ... + {x_n} - {x_i}}}} \ge \frac{{{{\left( {x_1^{5/2} + x_2^{5/2} + ... + x_n^{5/2}} \right)}^2}}}{{\left( {n - 1} \right)\left( {{x_1} + {x_2} + ... + {x_n}} \right)}}\] Cauchy-Schwarz again, we obtain \[\sqrt[5]{{{n^3}{{\left( {x_1^{5/2} + x_2^{5/2} + ... + x_n^{5/2}} \right)}^2}}} \ge {x_1} + {x_2} + ... + {x_n}\] , which implies: \[\frac{{{{\left( {x_1^{5/2} + x_2^{5/2} + ... + x_n^{5/2}} \right)}^2}}}{{\left( {n - 1} \right)\left( {{x_1} + {x_2} + ... + {x_n}} \right)}} \ge \frac{{{{\left( {x_1^{5/2} + x_2^{5/2} + ... + x_n^{5/2}} \right)}^{8/5}}}}{{\left( {n - 1} \right)\sqrt[5]{{{n^3}}}}}\] Applying Cauchy-Schwarz again, we obtain \[n{\left( {x_1^{5/2} + x_2^{5/2} + ... + x_n^{5/2}} \right)^4} \ge {\left( {x_1^2 + x_2^2 + ... + x_n^2} \right)^5} = 1\]\[ \Rightarrow x_1^{5/2} + x_2^{5/2} + ... + x_n^{5/2} \ge \frac{1}{{\sqrt[4]{n}}}\] So, we have \[\frac{{{{\left( {x_1^{5/2} + x_2^{5/2} + ... + x_n^{5/2}} \right)}^{8/5}}}}{{\left( {n - 1} \right)\sqrt[5]{{{n^3}}}}} \ge \frac{{{{\left( {\frac{1}{{\sqrt[4]{n}}}} \right)}^{8/5}}}}{{\left( {n - 1} \right)\sqrt[5]{{{n^3}}}}} = \frac{1}{{n\left( {n - 1} \right)}}\] Hence \[\sum\limits_{i = 1}^n {\frac{{x_i^5}}{{{x_1} + {x_2} + ... + {x_n} - {x_i}}}} \ge \frac{1}{{n\left( {n - 1} \right)}}\] Equality occurs if $a_1=a_2=...=a_n=\frac{1}{\sqrt{n}}$.

denote $N=x_1+\ldots+x_n$ and denote $\sum\limits_{i=1}^n\frac{x_i^5}{x_1+x_2+...+x_n-x_i}$ by $(X)$ by CS we have $(X)\sum\limits_{i=1}^n(x_i(N-x_i))\geq(x_1^3+\ldots+x_n^3)^2$ again CS $(x_1^3+\ldots+x_n^3)(N)\geq(x_1^2+\ldots+x_n^2)^2=1$ thus $(x_1^3+\dots+x_n^3)^2\geq\frac{1}{N^2}$ and we know that $\sum\limits_{i=1}^n(x_i(N-x_i))=N^2-1$ thus $(X)\geq\frac{(x_1^3+\ldots+x_n^3)^2}{\sum\limits_{i=1}^n(x_i(N-x_i))}\geq\frac{1}{N^4-N^2}$ we use CS again to show $(x_1+\ldots+x_n)\leq\sqrt{n(x_1^2+\ldots+x_n^2)}=\sqrt{n}$ from this we have $\frac{1}{N^4-N^2}\geq\frac{1}{n(n-1)}$ and finally from this follows $X\geq\frac{1}{n(n-1)}$ QED. i hope my solution is understandable

Holder just kills the problem !

please send your solution M-Nikdan!

Let $\sum_{i=1}^{n}\frac{x_{i}^{5}}{x_{1}+x_{2}+\ldots +x_{n}-x_{i}}=S$ Then by Holder's, \[S(\sum_{i=1}^{n} (\sum_{j=1}^{n} x_j-x_i))(\sum_{i=1}^{n} x_i) \ge (\sum_{i=1}^{n}x_i^2)^3\] Now rest should be easy

Notice that $\sum_{i}^{n}\frac{x_i^5}{x_1+x_2+\dots+x_n-x_i}\overset{\text{Generalized T2'S}}{\ge}\frac{\left(\sum_{i}^{n}x_i\right)^5}{n^3(n-1)\sum_{i}^{n}x_i}=\frac{\left(\sum_{i}^{n}x_i\right)^4}{n^3(n-1)}$ Claim: $\sum_{i}^nx_i\ge\frac{n}{\sqrt{n}}$ Proof: Let $f(x_1,x_2,\dots,x_n)=\sum_{i}^{n}x_i\text{ and }g(x_1,x_2,\dots,x_n)=\sum_{i}^nx_i^2-1\text{ and furthermore define }L=f(x_1,x_2,\dots,x_n)-\lambda g(x_1,x_2,\dots,x_n)$ $$\frac{\partial L}{\partial x_1}=1-2\lambda x_1=0\Longrightarrow x_1=\frac{1}{2\lambda}$$$$\vdots$$$$\frac{\partial L}{\partial x_n}=1-2\lambda x_n=0\Longrightarrow x_n=\frac{1}{2\lambda}$$Thus $x_1=\dots=x_n=\frac{1}{2\lambda}$, plugging this in yields $g(x_1,\dots,x_1)=nx_1^2-1=0\Longrightarrow x_1=\frac{1}{\sqrt{n}}$ therefore $x_1=\dots=x_n=\frac{1}{\sqrt{n}}$ So the minimum is obtained in the assertion $f\left(\frac{1}{\sqrt{n}},\dots,\frac{1}{\sqrt{n}}\right)=\frac{n}{\sqrt{n}}$ Using our last result we obtain that $\frac{\left(\sum_{i}^{n}x_i\right)^4}{n^3(n-1)}\ge\frac{n^2}{n^3(n-1)}=\frac{1}{n(n-1)}$ Thus the minimum of our original sum is $\sum_{i}^{n}\frac{x_i^5}{x_1+x_2+\dots+x_n-x_i}\ge\frac{1}{n(n-1)}$ $\blacksquare$.