hello, it is equivalent to $(x-1)^2(x^{n-1}+x^{n-2}+...+x+1-nx^{\frac{n-1}{2}})(x^{n-1}+x^{n-2}+...+x+1+nx^{\frac{n-1}{2}})\geq 0$ the second factor is AM-GM. Sonnhard.

We may assume $x\geq 1$, since otherwise it's the same thing for $x'= 1/x$. Take then $x=t^2$, with $t\geq 1$. It comes down to $t^n - 1/t^n \geq n(t-1/t)$, but $t^n - 1/t^n = (t-1/t)(t^{n-1} + t^{n-3} + \cdots + 1/t^{n-3} + 1/t^{n-1}) \geq n(t-1/t)$ by AM-GM. Equality only for $n=1$ and any $x>0$, or else for $n>1$ and $x=1$.

Sayan wrote: Prove that for every real number $x>0$ and each integer $n>0$ we have \[x^n+\frac1{x^n}-2 \ge n^2\left(x+\frac1x-2\right)\] Generalization Prove that for every real number $x>0$ and $n>m>0,n,m\in \mathbb{R}$ we have \[x^n+\frac1{x^n}-2 \ge \frac{{n}^2}{{m}^2}\left(x^m+\frac{1}{x^{m}}-2\right)\]

Sayan wrote: Prove that for every real number $x>0$ and each integer $n>0$ we have \[x^n+\frac1{x^n}-2 \ge n^2\left(x+\frac1x-2\right)\] particular case of inequality hence http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=491298

Let $x>0$ and $\alpha\geq1$ be real numbers. Prove that: \[ x^{\alpha}+\frac{1}{x^{\alpha}}-2\ge {\alpha}^2\left(x+\frac{1}x-2\right) \]

ionbursuc wrote: Sayan wrote: Prove that for every real number $x>0$ and each integer $n>0$ we have \[x^n+\frac1{x^n}-2 \ge n^2\left(x+\frac1x-2\right)\] Generalization Prove that for every real number $x>0$ and $n>m>0,n,m\in \mathbb{R}$ we have \[x^n+\frac1{x^n}-2 \ge \frac{{n}^2}{{m}^2}\left(x^m+\frac{1}{x^{m}}-2\right)\] $m=1,n=\alpha\Rightarrow $ Let $x>0$ and $\alpha\geq1$ be real numbers. Prove that: \[ x^{\alpha}+\frac{1}{x^{\alpha}}-2\ge {\alpha}^2\left(x+\frac{1}x-2\right) \]

OK, OK, OK! I missed your post.