# Prove inequality: When $n > 2$, $n! < {\left(\frac{n+2}{\sqrt{6}}\right)}^n$

Prove: When $n > 2$, $$n! < {\left(\frac{n+2}{\sqrt{6}}\right)}^n$$

PS: please do not use mathematical induction method.

EDIT: sorry, I forget another constraint, this problem should be solved by algebraic mean inequality.

Thanks.

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This used to be one of my favourite high-school problems. This is one approach: consider $y=\ln x$ and say that you want to integrate it between $1$ and $n$.

obviously the sum of the areas of trapezium $<\int_1^n\ln x\mathrm{d}x$. From this inequality, you get another inequality: $$n!<\left(\frac{n^{n+\frac{1}{2}}}{e^{n-1}}\right)$$ Then just show the following inequality and you are done: $$\left(\frac{n^{n+\frac{1}{2}}}{e^{n-1}}\right)<{\left(\frac{n+2}{\sqrt{6}}\right)}^n$$

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Stirlings approximation formula says

$$n! = \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n e^{\lambda_n}$$ where $$1/(12n+1)<\lambda_n<1/(12n)$$

Thus, it suffices to show $$\sqrt{2 \pi n} \left(\frac{n}{e}\right)^n e^{1/(12n)}<\left(\frac{n+2}{\sqrt{6}}\right)^n$$

This yields

$$\log(2 \pi n)/2 + n \log(n/e) + \frac{1}{12n} < n \log(n+2) - n \log(6)/2$$

and I think it should be fairly straightforward to prove this...

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I will use geometrical mean inequality. Suppose $$n!<\Bigg(\frac{n+2}{\sqrt{6}}\Bigg)^n\text{(for all n>2)}\rightarrow \sqrt{6}(n!)^{1/n}<(n+2)$$ is true. As $$(n!)^{1/n}\leq\frac{1+2+...+n}{n}=\frac{n(n+1)}{2n}=\frac{n+1}{2}\rightarrow$$ $$\rightarrow 2(n!)^{1/n}\leq n+1 \rightarrow 2(n!)^{1/n}<n+1 \text{ (for all n>1)}$$ Summing we have $$(2+\sqrt{6})(n!)^{1/n}<2n+3\text{ (for all n>2)}$$ then $$(n!)^{1/n}<\frac{2n+3}{2+\sqrt{6}}\text{(for all n>2)}$$ but $$(n!)^{1/n}<\frac{n+1}{2}\text{(for all n>2)}$$ You have to show $$\frac{n+1}{2}>\frac{2n+3}{2+\sqrt{6}}\text{(for all n>1+\sqrt{6})}$$ now we have to check cases $n=2$ and $3$ in the initial inequality, as initial conditions have changed as we calculated the difference between $2$ inequalities.