<s>[INST] 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.
 [/INST] 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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