Sequence $\left\{ a_n\right\}$ defined by $a_1=4$ , and $a_{n+1}=\sqrt{a_n+6}$. Prove that $\left| a_{n+1}-3\right|<\frac{1}{3}\left|a_n-3\right|$. As title. Sequence $\left\{a_n\right\}$ defined by $a_1=4$ , and $a_{n+1}=\sqrt{a_n+6}$. 
Prove that $\left| a_{n+1}-3\right|<\frac{1}{3}\left|a_n-3\right|$. 
I tried the induction. 
$a_{k+1}-3=\sqrt{a_k+6}-3<\sqrt{\frac{1}{3}\left(a_{k-1}-3\right)+6}-3=\frac{1}{3}\sqrt{a_{k-1}+15}-3$
But It seems not work. And I tryied trigonometrical substitution with letting $a_1=6\cos\theta$, then $a_2=\sqrt{6+6\cos\theta}=2\sqrt{3}\cos\frac{\theta}{2}$. And it seems not work either.
And I tried to square them to find a relation $\left(a_n-3\right)^2+6\left(a_n-3\right)=\left(a_{n-1}-3\right)$ or $\left(a_n-3\right)\left(a_n+3\right)=\left(a_{n-1}-3\right)$
It seems not work agian. Please help me, and thanks a lot.

If you define $b_n=a_n-3$, you want to show that $$|b_{n+1}|<\frac{|b_n|}{3}.$$
Now, the recursion is $$b_{n+1}=a_{n+1}-3=\sqrt{a_n+6}-3=\sqrt{b_n+9}-3=\frac{b_n}{\sqrt{b_n+9}+3}$$
Since $b_1=1$, this shows that $b_n>0$ for all $n$ by induction, so you can forgot about the absolute values. Moreover $\sqrt{b_n+9}>\sqrt{9}=3$, so $$\frac{1}{\sqrt{b_n+9}+3}<\frac 16,$$
hence $$b_{n+1}=\frac{b_n}{\sqrt{b_n+9}+3}<\frac{b_n}{6}$$
which in fact it is stronger to what you need. 
