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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.

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  • $\begingroup$ Note: equality holds when $n=6$ $\endgroup$ – J. W. Tanner Jul 9 at 17:40
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    $\begingroup$ You'd have to ask them how they saw it. We can only answer how we can see it. $\endgroup$ – Robert Israel Jul 9 at 17:44
  • $\begingroup$ One tempting method is to write this as $f(n) < g(n)$ and then show that $\log (f(n)/g(n))$ is a decreasing function of $n$ for $n > 6$. However this method won't work because that function isn't decreasing. $\endgroup$ – Michael Lugo Jul 9 at 18:31
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Raise both sides to the $n$th power: $$ \left(\frac{n+2}{n-2}\right)^n < (n+2)^2, $$ then multiply through by $(n-2)^2/(n+2)^2$: $$ \left(\frac{n+2}{n-2}\right)^{n-2} = \left(1+\frac{4}{n-2}\right)^{n-2} < (n-2)^2. $$ The left side is bounded by $e^4$, so as long as $n-2 > e^2$, the inequality is guaranteed to be satisfied. This proves it true for all $n \ge 10$, and the remaining cases can be checked individually.

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Raising both sides to the $n$-th power, we obtain:

$$\left(\dfrac{n+2}{n-2}\right)^n<(n+2)^2,$$

which is valid since $n>1$. I note that now that RHS is a concave up quadractic in $n$, while in the limit at infinity, the LHS will be $1$ (Note, we could quibble about whether that last phrase is rigorous, but it is certainly true in the original expression of the claim, before we raise to the $n$-th power). To verify the inequality, then, is to note the following:

1) For $n>2$ The LHS is a decreasing function and the RHS is an increasing function.

2) They are equal when $n=6$, as was pointed out in the comments.

Therefore, for $n>6$, or $n\geq 7$ in our integer context here, the claim is verified.

As far as your question of how the authors noticed this, I can only speculate, but I would guess that this came across as a consequence of something they wanted to prove, and they went about proving it. Just speculation, but that is often how such things happen for me.

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For all $n\geq11$ $$\left(\frac{n+2}{n-2}\right)^n=\left(1+\frac{4}{n-2}\right)^{\frac{n-2}{4}\cdot\frac{4n}{n-2}}<e^{\frac{4n}{n-2}}\leq e^{\frac{44}{9}}<(n+2)^2.$$ Thus, it's enough to check $n\in\{7,8,9,10\}$.

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(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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Depending on what you're doing, you either get the functions $$ f(n):=\frac{n+2}{n-2}\qquad,\qquad g(n):=(n+2)^{2/n} $$ out of your calculations, or you discover one of them in the process of trying to find suitable formula for your proof/calculation (the other should be already given, because otherwise, what are you even doing?).

In the second case, you first start at a formula with lots of variables, that has a general form that you think might help you get further.
Then you start trying out a few parameters, try to get a feel if this function can actually fit what you want.

If so, you proceed further, to the place you'd start in the first case. To show inequalities for all $x\in \mathbb R, x>c$ for some constant $c$, there are two useful ways (that I can think of at the top of my head):

#1: If the formulas are both differentiable, it is sufficient to show that one of them grows faster than the other from some point on, or more specifically:
If there exists a $C\in\mathbb R$, so that $\forall x\in\mathbb: f^{(i)}(x)>g^{(i)}(x)$ for some $i\in\mathbb N$, then at some point $c$ we have $\forall x\in\mathbb R: f(x)>g(x)$.

#2: If the formulas are continuous, we know that between two adjacent intersection points $x_1,x_2$ for the whole interval $(x_1,x_2)$ either $f(x)<g(x)$ or $f(x)>g(x)$.
This also holds for the interval of the last intersection point till infinity, or from negative infinity to the first intersection point.

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