<s>[INST] 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.
 [/INST] Hint :
$$
a_{n+1}-3=\sqrt{a_n+6}-\sqrt{9}=\frac{a_n+6-9}{\sqrt{a_n+6}+\sqrt{9}}
$$
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