How prove that: $[12\sqrt[n]{n!}]{\leq}7n+5$? How prove that: $[12\sqrt[n]{n!}]{\leq}7n+5$,$n\in N$  I know $\lim_{n\to  \infty  } (1+ \frac{7}{7n+5} )^{ n+1}=e$ and $\lim_{n\to  \infty  }  \sqrt[n+1]{n+1} =1$.
 A: By AM-GM 
$$\frac{1+2 + 3 + \cdots + n}{n} \ge \sqrt[n]{1 \times 2 \times 3 \times \cdots \times n}$$
$$\implies \frac{n+1}2 \ge \sqrt[n]{n!} \implies 6n+6 \ge 12\sqrt[n]{n!}$$
But $7n+5 \ge 6n+6$ for $n \ge 1$...
A: I'm assuming that $\left[x\right]
 $ is the floor function. For $n=1,2
 $ works. So assume that $n\geq3
 $. Using the bound $$n!\leq\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}e^{1/\left(12n\right)}
 $$ we have $$\left[12\sqrt[n]{n!}\right]\leq12\sqrt[n]{n!}\leq12\left(2\pi n\right)^{1/\left(2n\right)}e^{1/\left(12n^{2}\right)-1}n
 $$ so we have to prove now that $$12\left(2\pi n\right)^{1/\left(2n\right)}e^{1/\left(12n^{2}\right)-1}n\leq7n+5
 $$ and we can prove it by induction. For $n=3
 $ works, so we consider 
\begin{align*}
12\left(2\pi\left(n+1\right)\right)^{1/\left(2\left(n+1\right)\right)}e^{1/\left(12\left(n+1\right)^{2}\right)-1}\left(n+1\right)= & 12\left(2\pi\left(n+1\right)\right)^{1/\left(2\left(n+1\right)\right)}e^{1/\left(12\left(n+1\right)^{2}\right)-1}n\\
+ & 12\left(2\pi\left(n+1\right)\right)^{1/\left(2\left(n+1\right)\right)}e^{1/\left(12\left(n+1\right)^{2}\right)-1}\tag{1}
\end{align*}
and since $12\left(2\pi x\right)^{1/\left(2x\right)}e^{1/\left(12x^{2}\right)-1}$ is a monotone decreasing function for $x\geq1
 $ we observe that $$
12\left(2\pi\left(n+1\right)\right)^{1/\left(2\left(n+1\right)\right)}e^{1/\left(12\left(n+1\right)^{2}\right)-1}n\leq12\left(2\pi n\right)^{1/\left(2n\right)}e^{1/\left(12n^{2}\right)-1}n
 $$ hence $$
(1)\leq7n+5+12\left(2\pi\left(n+1\right)\right)^{1/\left(2\left(n+1\right)\right)}e^{1/\left(12\left(n+1\right)^{2}\right)-1}\leq7\left(n+1\right)+5
 $$ since $$12\left(2\pi\left(n+1\right)\right)^{1/\left(2\left(n+1\right)\right)}e^{1/\left(12\left(n+1\right)^{2}\right)-1}\leq12\left(8\pi\right)^{1/8}e^{1/192-1}<7.
 $$
