# 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.

• Note: equality holds when $n=6$ – J. W. Tanner Jul 9 at 17:40
• You'd have to ask them how they saw it. We can only answer how we can see it. – Robert Israel Jul 9 at 17:44
• 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. – Michael Lugo Jul 9 at 18:31

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.

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.

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

(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 $$x \geq 7$$.

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 $$f(x) or $$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.