Silk Road Mathematics Competition

1) By the rearrangement inequality we have: $$\begin{array}{rl}{a}_1^m+a_2^m+\cdots +a_n^m& \ge a_1^p{a}_2^k+a_2^p{a}_3^k+\cdots +a_{n-1}^p{a}_n^k+a_n^p{a}_1^k\\ a_1^m+a_2^m+\cdots +a_n^m& \ge a_1^p{a}_3^k+a_2^p{a}_4^k+\cdots +a_{n-1}^p{a}_1^k+a_n^p{a}_2^k\\ \text{multicolumn}2c\text{...................}& \text{...................}\\ a_1^m+a_2^m+\cdots +a_n^m& \ge a_1^p{a}_t^k+a_2^p{a}_{t+1}^k+\cdots +a_{n-1}^p{a}_{t-2}^k+a_n^p{a}_{t-1}^k\end{array}$$ Adding above $(t-1)$ inequalities we get $$(1)(t-1)(a_1^m+a_2^m+\cdots +a_n^m)\ge a_1^p(a_2^k+a_3^k+\cdots +a_t^k)+a_2^p(a_3^k+a_4^k+\cdots +a_{t+1}^k)+\cdots +a_n^p(a_1^k+a_2^k+\cdots +a_{t-1}^k)$$

and using Cauchy-Schwartz inequality we have $$(t-1)(a_1^m+a_2^m+\cdots +a_n^m)\left(\frac{a_1^p}{a_2^k+a_3^k+\cdots +a_t^k}+\frac{a_2^p}{a_3^k+a_4^k+\cdots +a_{t+1}^k}+\cdots +\frac{a_n^p}{a_1^k+a_2^k+\cdots +a_{t-1}^k}\right)\ge (a_1^p+a_2^p+\cdots +a_n^p{)}^2$$ Now, our conclusion is obtained by dividing both sides of above equation by $(t-1)(a_1^m+a_2^m+\cdots +a_n^m)$.

2) Multiplying both sides of (1) by $$\frac{a_2^k+a_3^k+\cdots +a_t^k}{a_1^p}+\frac{a_3^k+a_4^k+\cdots +a_{t+1}^k}{a_2^p}+\cdots +\frac{a_1^k+a_2^k+\cdots +a_{t-1}^k}{a_n^p}$$ and using Cauchy-Schwartz inequality too we have $$(t-1)(a_1^m+a_2^m+\cdots +a_n^m)\left(\frac{a_2^k+a_3^k+\cdots +a_t^k}{a_1^p}+\frac{a_3^k+a_4^k+\cdots +a_{t+1}^k}{a_2^p}+\cdots +\frac{a_1^k+a_2^k+\cdots +a_{t-1}^k}{a_n^p}\right)\ge [(t-1)(a_1^k+a_2^k+\cdots +a_n^k){]}^2$$ Again, our conclusion is obtained by dividing both sides of above equation by $(t-1)(a_1^m+a_2^m+\cdots +a_n^m)$.