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I am not sure whether the following inequality is true? Some small $n$ indicates it is true.

Let $n$ be a positive integer and $c\gt0$, then $$c^n(n!+c^n)\lt(n+c^2)^n.$$

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3 Answers 3

up vote 5 down vote accepted

Consider $n = 1$ and $c = 10$. Then, $c^n(n!+c^n) = 110$ while $(n+c^2)^n = 101$. You will be able to show a number of other similar counterexamples when $n=1$.

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+1: Some small $n$ indicates it is true is not really true then :-) –  Aryabhata May 14 '11 at 21:21
    
Sorry guys, I made a wrong formulation. I will update it soon. Can anyone help delete this post? –  Sunni May 14 '11 at 21:32

I believe this is false.

Take $n=25, c = 5$.

I came up with this example by setting $c = \sqrt{n}$ and using Stirling's approximation formula for $n!$.

Wolfram alpha link showing the computation.

In fact for $c = \sqrt{n}$, we have that

$$\frac{c^n(n!+c^n)}{(n+c^2)^n} \sim \sqrt{2 \pi n}\ \left(\frac{n}{\sqrt{2e}}\right)^{n/2}$$

which goes to $\infty$ as $n \to \infty$.

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Wow, you don't like trivial cases as $n=1$, do you. XD –  Patrick Da Silva May 15 '11 at 7:25
    
@Patrick: I actually ignored those, as OP said those seem to work... –  Aryabhata May 15 '11 at 14:28

For $n=1$, you have $c+c^2 < 1+c^2$, which is equivalent to $c < 1$, hence always false because you supposed $c > 0$.

For $n=2$, you have $2c^2 + c^4 < c^4 +4c^2 + 4$, which is equivalent to $2c^2 > -4$ (always true).

Aryabhatta's example ensures us that we cannot conjecture that the inequality will always be true for $n > 2$.

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