Prove that if $x_1, …, x_{n-1}$ are positive numbers and $n \geq 2$, than the following inequality holds:
$x_1^2 + x_2^3 + … + x_{n – 1}^n + \frac{1}{x_1^2 x_2^3 … x_{n – 1}^n} \geq n + (x_1 – 1)^2 + 2(x_2 – 1)^2 + … + (n – 1)(x_{n – 1} – 1)^2$.
Firstly, I tried to prove this inequality by induction, but it didn't work. Then I tried using the AM-GM inequality, and finally some substitutions, but none of them lead me to a nicer form.
Do you have any suggestions for this inequality?
Best Answer
For all $n\geq2$ we need to prove that $$\sum_{k=1}^{n-1}(x_k^{k+1}-k(x_k-1)^2)+\frac{1}{\prod\limits_{k=1}^{n-1}x_k^{k+1}}\geq n$$ or $$\sum_{k=1}^{n-1}(x_k^{k+1}-kx_k^2+2kx_k)+\frac{1}{\prod\limits_{k=1}^{n-1}x_k^{k+1}}\geq\frac{n(n-1)}{2}+n$$ or $$\sum_{k=1}^{n-1}(x_k^{k+1}-kx_k^2+(k-1)x_k)+\sum_{k=1}^{n-1}(k+1)x_k+\frac{1}{\prod\limits_{k=1}^{n-1}x_k^{k+1}}\geq\frac{n(n+1)}{2},$$ which is true by AM-GM: $$x_k^{k+1}-kx_k^2+(k-1)x_k=k_k(x_k^k+(k-1)\cdot1-kx_k)\geq0$$ and $$\sum_{k=1}^{n-1}(k+1)x_k+\frac{1}{\prod\limits_{k=1}^{n-1}x_k^{k+1}}\geq\frac{n(n+1)}{2}\left(\prod\limits_{k=1}^{n-1}x_k^{k+1}\cdot\frac{1}{\prod\limits_{k=1}^{n-1}x_k^{k+1}}\right)^{\frac{2}{n(n+1)}}=\frac{n(n+1)}{2}.$$