Find the limit of a recursive sequence Let $(u_n)_n$ be a real sequence such that
$$
u_{n+2}=\sqrt{u_{n+1}}+\sqrt{u_{n}},\,u_0>0,\,u_1>0.
$$
Fisrt, it is easy to check that $(u_n)_n$ is well defined and $u_n>0$ for all $n\in\mathbb{N}$.
The question now is show that 
$$
\exists p\in \mathbb{N}\,;\,\forall n\in\mathbb{N},\,n\geq p\implies u_n>1.
$$
From this, we can deduce the limit of the sequence $(u_n)_n$.
 A: Lemma.  If $c$ is a real number and $0<c<\frac12$, then the quadratic $q(\lambda)=\lambda^2-c\lambda-c$ has a real root $\alpha$ strictly between $0$ and $1$.
Proof: $q(0)=-c<0$ and $q(1)=1-2c>0$.

Lemma.  Let $c$ be real, $0<c<\frac12$.  If a sequence of non-negative numbers satisfies
$$a_n\le c(a_{n-1}+a_{n-2})$$
for all $n\ge0$, then there is a non-negative constant $A$ such that
$$a_n\le A\alpha^n$$
for all $n$, where $\alpha$ is the constant in the previous lemma.

Proof.  Choose
$$A=\max\Bigl(a_0,\frac{a_1}\alpha\Bigr)\ .$$
Then the inequality is clearly true for $n=0,1$; and since $c\alpha+c=\alpha^2$, the result follows easily by induction.

Lemma.  Let $\{u_n\}$ be the sequence defined in the question.  Then
$$u_n\ge\min(1,u_1)$$
for all $n\ge1$.

Proof.  If $u_{n-1}<1$ then
$$u_n=\sqrt{u_{n-1}}+\sqrt{u_{n-2}}>\sqrt{u_{n-1}}>u_{n-1}\ ,$$
while if $u_{n-1}\ge1$ then
$$u_n=\sqrt{u_{n-1}}+\sqrt{u_{n-2}}>\sqrt{u_{n-1}}>1\ .$$
The result follows by induction.

Solution of the problem.  We have
$$\eqalign{|u_n-4|
  &=\bigl|(\sqrt{u_{n-1}}-2)+(\sqrt{u_{n-2}}-2)\bigr|\cr
  &\le\bigl|\sqrt{u_{n-1}}-2\bigr|+\bigl|\sqrt{u_{n-2}}-2\bigr|\cr
  &=\left|\frac{u_{n-1}-4}{\sqrt{u_{n-1}}+2}\right|
    +\left|\frac{u_{n-2}-4}{\sqrt{u_{n-2}}+2}\right|\cr
  &\le c\bigl(|u_{n-1}-4|+|u_{n-2}-4|\bigr)\cr}$$
where
$$c=\frac{1}{2+\sqrt{\min(1,u_1)}}<\frac12\ .$$
It follows from the lemmas that $|u_n-4|\to0$ as $n\to\infty$.
