Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

orl wrote: Prove the following inequality: \[\prod^k_{i=1} x_i \cdot \sum^k_{i=1} x^{n-1}_i \leq \sum^k_{i=1} x^{n+k-1}_i,\] where $x_i > 0,$ $k \in \mathbb{N}, n \in \mathbb{N}.$ This is a direct result of Muirhead's inequality; $(n+k-1,0,0,...0)$ majorizes $(n-1,1,1,...1)$. Alternatively, it follows from the AM-GM by the symmetric sum of $nx_1^{n+k-1}+x_2^{n+k-1}+...+x_k^{n+k-1} \geq (n+k-1)x_1^{n}x_2...x_k$.

I think you mean (n+k-1,0,0,...,0)\succ(n,1,1,...,1)

Or we can use Chebishev Inequality: for sequences $ \left\{a_{n}\right\}$ and $ \left\{b_{n}\right\}$ that both are increasing or both decreasing, and real numbers $ \mu_{i}\;\;(1\leq i\leq n)$ we have: $ \boxed{\sum^{n}_{i = 1}\mu_{i}a_{i}\sum^{n}_{i = 1}\mu_{i}b_{i}\leq\sum^{n}_{i = 1}a_{i}b_{i}\sum^{n}_{i = 1}\mu_{i}}$ now we have: $ \sum^{k}_{i = 1}(1\cdot x^{k}_{i})\sum^{k}_{i = 1}(1\cdot x^{n-1}_{i})\leq\sum^{k}_{i = 1}(x^{k}_{i}\cdot x^{n-1}_{i})\sum^{k}_{i = 1}1$ $ \Longrightarrow\frac{1}{k}\sum^{k}_{i = 1}x^{k}_{i}\cdot\sum^{k}_{i = 1}x^{n-1}_{i}\leq\sum^{k}_{i = 1}x^{n+k-1}_{i}$ by AM-GM : $ \prod^{k}_{i = 1}x_{i}\leq\frac{1}{k}\sum^{k}_{i = 1}x^{k}_{i}$ and the desired inequality is obtained!