Modified rearrangement inequality..probably.. Prove that for real numbers $x_{n+1}=x_1\ge x_2\ge ...\ge x_n\ge 0$,$$\sum_{k=1}^n\frac {x_k}{x_{k+1}} \le \sum_{k=1}^n\frac {x_{k+1}}{x_k}$$Bringing LHS to RHS, we get $$\sum_{k=1}^n\frac {x_{k+1}^2-x_k^2}{x_k.x_{k+1}}\ge 0$$ Let $$T=\sum_{k=1}^n\frac {x_{k+1}^2-x_k^2}{x_k.x_{k+1}}$$ and $$S=\sum_{k=1}^n\frac {x_{k}^2+x_{k+1}^2}{x_k.x_{k+1}}$$

The inequality follows by induction. For $n=2$ we have an equality. Now,
suppose that if $x_{1}\geq x_{2}\geq\ldots\geq x_{n}\geq0$, then
$$
\frac{x_{1}}{x_{2}}+\ldots+\frac{x_{n-1}}{x_{n}}+\frac{x_{n}}{x_{1}}\leq
\frac{x_{2}}{x_{1}}+\ldots+\frac{x_{n}}{x_{n-1}}+\frac{x_{1}}{x_{n}},
$$
and let $x_{1}\geq x_{2}\geq\ldots\geq x_{n}\geq x_{n+1}\geq0.$ Observe that
$$
\frac{x_{n}}{x_{n+1}}+\frac{x_{n+1}}{x_{1}}-\frac{x_{n}}{x_{1}}\leq
\frac{x_{n+1}}{x_{n}}+\frac{x_{1}}{x_{n+1}}-\frac{x_{1}}{x_{n}},
$$
since denoting $\frac{x_{n+1}}{x_{n}}=a\leq1,$ and $\frac{x_{n}}{x_{1}}
=b\leq1,$ the latter simplifies to
$$
\left(  1-ab\right)  \left(  1-a\right)  \left(  1-b\right)  \geq0.
$$
We obtain
$$
\frac{x_{1}}{x_{2}}+\ldots+\frac{x_{n}}{x_{n+1}}+\frac{x_{n+1}}{x_{1}}
\leq\frac{x_{2}}{x_{1}}+\ldots+\frac{x_{n+1}}{x_{n}}+\frac{x_{1}}{x_{n+1}},
$$
as desired.
