# If $x_1\geq x_2\geq…\geq x_n\geq0,$ $n\geq2$, $m\geq k\geq1$ then $x_1^mx_2^k+x_2^mx_3^k+…+x_n^mx_1^k\geq x_1^kx_2^m+x_2^kx_3^m+…+x_n^kx_1^m.$

Is the following inequality is true??

If $$x_1\geq x_2\geq...\geq x_n\geq0,$$ $$n\geq2$$, $$m\geq k\geq1$$ then $$x_1^mx_2^k+x_2^mx_3^k+...+x_n^mx_1^k\geq x_1^kx_2^m+x_2^kx_3^m+...+x_n^kx_1^m.$$

My answer:Let's try for 3 variables. Let $$m=k+r.$$ $$a^m b^k + b^m c^k + c^m a^k - (a^k b^m + b^k c^m + c^k a^m)$$ $$=a^k b^k (a^r - b^r) +b^k c^k (b^r - c^r) +c^k a^k (c^r - a^r)$$ $$=a^k b^k (a^r - b^r) +b^k c^k (b^r - c^r) +c^k a^k (c^r - b^r + b^r -a^r)$$ $$=a^k (b^k - c^k)(a^r - b^r) - c^k (a^k - b^k)(b^r - c^r)$$ I hope this is always non-negative, but can't see any neat proof!Please help me.

Because for three variables it's true (a nice exercise on Mean value theorem) and for $$n\geq3$$ we obtain: $$\sum_{i=1}^n\left(x_i^mx_{i+1}^k-x_i^kx_{k+1}^m\right)=\sum_{i=1}^n(x_ix_{i+1})^k(x_i^r-x_{i+1}^r)=$$ $$=\sum_{i=2}^{n-1}\left((x_1x_i)^k(x_1^r-x_{i}^r)+(x_{i}x_{i+1})^k(x_{i}^r-x_{i+1}^r)+(x_{i+1}x_1)^k(x_{i+1}^r-x_{1}^r)\right)\geq0.$$
For example, for five variables $$a\geq b\geq c\geq d\geq e\geq0$$ it looks so: $$(ab)^k(a^r-b^r)+(bc)^k(b^r-c^r)+(cd)^k(c^r-d^r)+(de)^k(d^r-e^r)+(ea)^k(e^r-a^r)=$$ $$=(ab)^k(a^r-b^r)+(bc)^k(b^r-c^r)+(ca)^k(c^r-a^r)+$$ $$+(ac)^k(a^r-c^r)+(cd)^k(c^r-d^r)+(da)^k(d^r-a^r)+$$ $$+(ad)^k(a^r-d^r)+(de)^k(d^r-e^r)+(ea)^k(e^r-a^r)\geq0.$$