Prove Some questions with mathematical induction I have some question and I want to prove them with mathematical induction.  
1) $|\sin(nx)| \le n|\sin(x)|$ 
2) $\sqrt[n]{n!} \ge \sqrt{n}$ 
3) $1\cdot3\cdot5\cdots(2n-1) \le n^n$  
I use this method for question one but can you find easier way?
$P(0) = |\sin(0)| \le 0*|\sin(x)|$ It's true.  
$P(n) = |\sin(nx)| \le n|\sin(x)|$  
$P(n+1) = |\sin(nx + x)| \le (n+1)|\sin(x)|$   
$\sin(nx + x)| = |\sin(nx)\cos(x) + \cos(nx)sin(x)|$  
$|\sin(nx)\cos(x) + \cos(nx)sin(x)| \le |\sin(nx)\cos(x)| + |\cos(nx)sin(x)|$  
$|\sin(nx)\cos(x)| + |\cos(nx)sin(x)| \le |n\sin(x)\cos(x)| + |\cos(nx)sin(x)|$ (we use p(n) for this)  
$|n\sin(x)\cos(x)| + |\cos(nx)sin(x)| = |\sin(x)|(n|\cos(x)| + |\cos(nx)|)$  
now, we must prove that:
$|\sin(x)|(n|\cos(x)| + |\cos(nx)|) \le (n+1)|\sin(x)| $ or same meaning
$(n|\cos(x)| + |\cos(nx)| \le n+1$   
It's obvious because of domain of cos(x).
Is it possible to help me?
Thanks.
 A: HINTS:
HINT (i): Using the triangle inequality, we have
$$\begin{align}
|\sin((n+1)x)|&=|\sin(nx)\cos(x)+\cos(nx)\sin(x)|\\\\
&\le |\sin(nx)|+|\sin(x)| \tag 1
\end{align}$$

HINT (ii):  The inequality is equivalent to showing
$$n!\ge n^{n/2}$$
Then, note that
$$\begin{align}
(n+1)^{(n+1)/2}&=n^{n/2}(n+1)^{1/2}\left(1+\frac1n\right)^{n/2}\\\\
&\le n!(n+1)^{1/2}\left(1+\frac1n\right)^{n/2} \,\,\cdots \text{by the induction hypothesis} \tag 2
\end{align}$$
Use the fact that $\left(1+\frac1n\right)^{n/2}= \sqrt{2}$ for $n=1$ and $\left(1+\frac1n\right)^{n/2}\le e^{1/2}<\sqrt{3}$ for all $n$.

HINT (iii):  Write $(2n-1)!!=\frac{(2n)!!}{2^n\,n!}$.  Then, the inequality is equivalent to showing 
$$(2n)!!\le (2n)^n\,n!$$
Note that we have
$$\begin{align}
(2n+2)!!&=2(n+1)(2n)!!\\\\
&\le 2(n+1)(2n)^n\,n!\,\,\cdots \text{by the induction hypothesis} \tag 3
\end{align}$$

SPOILER ALERT:  Scroll over the highlighted area to reveal the solutions.

From $(1)$ and the induction hypothesis $|\sin(nx)|\le n|\sin(x)|$, we see that $$\begin{align}|\sin((n+1)x)|&\le n|\sin(x)|+|\sin(x)|\\\\&=(n+1)|\sin(x)|\end{align}$$And we are done! From $(2)$, we have $$\begin{align}(n+1)^{(n+1)/2}&\le (n+1)!\,\frac{\left(1+\frac1n\right)^{n/2}}{(n+1)^{1/2}}\\\\&\le (n+1)!\end{align}$$since for $n=1$, $\frac{\left(1+\frac1n\right)^{n/2}}{(n+1)^{1/2}}=1$ and for $n\ge 2$, $\frac{\left(1+\frac1n\right)^{n/2}}{(n+1)^{1/2}}\le \sqrt{e/(n+1)}<1$.  And we are done!  From $(3)$, we see that $$\begin{align}(2n+2)!!&\le  2(n+1)(2n)^n\,n!\\\\&=(2n+2)^{n+1}(n+1)!\left(\frac{n^n}{(n+1)^{n+1}}\right)\\\\&<(2n+2)^{n+1}(n+1)!\end{align}$$And we are done!

