find the maximum $\frac{\frac{x^2_{1}}{x_{2}}+\frac{x^2_{2}}{x_{3}}+\cdots+\frac{x^2_{n-1}}{x_{n}}+\frac{x^2_{n}}{x_{1}}}{x_{1}+x_{2}+\cdots+x_{n}}$ give the postive intger $n\ge 2$,and postive real numbers $a<b$ if the real numbers such $x_{1},x_{2},\cdots,x_{n}\in[a,b]$ find the maximum of the value
$$\dfrac{\frac{x^2_{1}}{x_{2}}+\frac{x^2_{2}}{x_{3}}+\cdots+\frac{x^2_{n-1}}{x_{n}}+\frac{x^2_{n}}{x_{1}}}{x_{1}+x_{2}+\cdots+x_{n}}$$
it seem the  polya-szego inequality   http://journalofinequalitiesandapplications.springeropen.com/articles/10.1186/1029-242X-2013-591

Let $M$ is a maximum value (it exists because continuous function on compact gets there a maximum value) and
$$f(x_1,x_2,...x_n)=\frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+...+\frac{x_n^2}{x_1}-M(x_1+x_2+...+x_n).$$
Since  $f$ is a convex function for all $x_i$, we obtain
$$0=\max_{\{x_1,x_2,...,x_n\}\subset[a,b]}f=\max_{\{x_1,x_2,...,x_n\}\subset\{a,b\}}f$$
From here if $n$ is even we have $$M=\frac{\frac{a^2}{b}+\frac{b^2}{a}}{a+b}=\frac{a}{b}+\frac{b}{a}-1,$$
which occurs for $x_1=x_3=...=a$ and $x_2=x_4=...=b$.
If $n$ is odd we have for $n=2m+1$:
$$M=\frac{m\left(\frac{a^2}{b}+\frac{b^2}{a}\right)+a}{(m+1)a+mb},$$
which occurs for $x_1=x_3=...=x_{2m+1}=a$ and $x_2=x_4=...=x_{2m}=b$.
