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 $\lim_{n\to  \infty  }  \sqrt[n+1]{n+1} =1$.

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 $n \ge 1$...
