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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 $n$.

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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 $$1/(12n+1)<\lambda_n<1/(12n)$$

see http://en.wikipedia.org/wiki/Stirling%27s_approximation

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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For $a_1, a_2, \ldots, a_n \geqslant 0$, we have:$$\sqrt [ n ]{ { a }_{ 1 }\cdot { a }_{ 2 }\cdots { a }_{ n } } \leqslant \frac { { a }_{ 1 }+{ a }_{ 2 }+\cdots +{ a }_{ n } }{ n } $$

Consider ${ \left( n! \right) }^{ 2 }=\left( n\cdot 1 \right) \left( \left( n-1 \right) \cdot 2 \right) \cdots \left( 1\cdot n \right) $, for $n>1$:

$$\sqrt [ n ]{ { \left( n! \right) }^{ 2 } } =\sqrt [ n ]{ \left( n\cdot 1 \right) \left( \left( n-1 \right) \cdot 2 \right) \cdots \left( 1\cdot n \right) } \leqslant \frac { n\cdot 1+\left( n-1 \right) \cdot 2+\cdots +\left( n-\left( n-1 \right) \right) \cdot n }{ n } \\ =\frac { n\left( 1+2+\cdots +n \right) -1\times 2-2\times 3-\cdots -\left( n-1 \right) \times n }{ n } \\ =\frac { n\left( \frac { n\left( n+1 \right) }{ 2 } \right) -\frac { n\left( { n }^{ 2 }-1 \right) }{ 3 } }{ n } \\ =\frac { \left( n+1 \right) \left( n+2 \right) }{ 6 } <\frac { { \left( n+2 \right) }^{ 2 } }{ 6 } $$

That's the conclusion, which we can get using high school math not the Stirlings.

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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.

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