<s>[INST] Why $ \frac{n+2}{n-2}<(n+2)^{2/n}$ for $n\geq 7$. In some paper, the authors mentioned the following statement: 
One can easily check that for $n\geq  7$, 
$$ \frac{n+2}{n-2}<(n+2)^{2/n}.$$
This statement is correct, and their objective was to find an upper bound of $\frac{n+2}{n-2}$, eventually starting from some integer. Now my question is how we can see that $(n+2)^{2/n}$ is an upper bound for $\frac{n+2}{n-2}$ starting from some integer ( here it is $7$).
Thank you. 
 [/INST] (1) The inequality is true for $n=7$.
(2) Assume $(k+2)^{\frac{k-2}{k}} < k - 2$
(3) We must show $(k+3)^{\frac{k-1}{k+1}} < k - 1$. 
Consider $f(x) = (x+3)^{\frac{x-1}{x+1}} - x + 1$. (We want to show $f(x) < 0$ for $x \geq 7$.) 
Well, $(x+3)^{\frac{x-1}{x+1}} - x + 1 < 0$ iff $(x+3)^{\frac{x-1}{x+1}} < x - 1$ iff $\frac{x-1}{x+1}\ln(x+3) < \ln(x-1)$ iff $\frac{x-1}{x+1} < \frac{\ln(x-1)}{\ln(x+3)},$ which is true for $x=7$; and, since $\frac{d}{dx}\left(\frac{x- 1}{x+1}\right) < 1$ and $\frac{d}{dx}\left(\frac{\ln(x-1)}{\ln(x+3)}\right) > 1$ for all $x > 7$, the last inequality holds for all $x \geq 7$. 
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