Math Help - Sequence

1. Sequence

Hi,

$(\forall n \in \mathbb{N}) U_{n+1}=\sqrt{12+U_n}$ and $\ U_0=0$

I showed that $(\forall n \in \mathbb{N}) U_n<4$

But i don't know how to show that $4-U_{n+1}\le \frac{1}{4}(4-U_n)$

Can you help me please????

2. 4-U(n+1)=4-sqrt(12+U(n)),
using a=4-U(n)
4-sqrt(16-a)=
=4-4*sqrt(1-a/16) (1)
>=4-4(1-(1/2)a/16)
4-U(n+1)>=(1/8)(4-U(n)) - aditional rule.
Using Taulor set of sqrt:
sqrt(1-x)>=1-x and multiplying by -1, we get
-sqrt(1-x)<=-(1-x)
from (1) we get
4-U(n+1)<=(1/4)(4-U(n)).