 Solution

Let $F(x)=\sqrt{2+\sqrt{x}}$ for $x\ge 0$. This is an increasing continuous function. Starting at $a_1=\sqrt{2}$, induction gives $\sqrt{2}\le a_n\le 2$: if $a_n\le 2$, then $a_{n+1}\le \sqrt{2+\sqrt{2}}<2$, and every iterate is at least $\sqrt{2}$.

The first step has $a_2>\sqrt{2}=a_1$. Since $F$ is increasing, $a_n\ge a_{n-1}$ implies $a_{n+1}=F(a_n)\ge F(a_{n-1})=a_n$. Thus this sequence is increasing and bounded above. Let $L_{-}={\sup }_n{a}_n$. For every $\epsilon >0$, a term exceeds $L_{-}-\epsilon$; all later terms lie between it and $L_{-}$. This proves convergence, the elementary bounded monotone sequence theorem.

For the second starting value, write the iterates as $b_1=4$ and $b_{n+1}=F(b_n)$. Here $b_2=2<4$. Monotonicity of $F$ propagates this inequality, so $b_n$ is decreasing. It is bounded below by $\sqrt{2}$ because every positive iterate is at least that value. Hence it converges to its infimum $L_{+}$. From $b_2=2$ onward all terms lie in $[\sqrt{2},2]$.

Continuity of $F$ permits passing to each limit in the recurrence, giving $L_{\pm }^2=2+\sqrt{L_{\pm }}$. To prove that these fixed points agree, suppose $r,s\ge \sqrt{2}$ both satisfy that equality. Subtracting the two equations and rationalizing givesThe second factor is positive: $r+s\ge 2\sqrt{2}$, whereas $(\sqrt{r}+\sqrt{s}{)}^{-1}\le 1/(2\sqrt{\sqrt{2}})<2\sqrt{2}$. Thus $r=s$. ThereforeExistence has been supplied by the two convergent sequences, and the subtraction argument gives the required uniqueness in their common interval.