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The problem is:

Define a sequence by $a_{1} = 1, a_{2} = 1, a_{n+2} = \sqrt{a_{n+1} + a_{n}}, \forall n \geq 1.$

(a) Prove that $a_{n} < 2$, for all positive integer $n$.

(b) Prove that for all positive integer $n$ such that $n \geq 2$, we have $a_{n+1} > a_{n}$.

I have trouble with part (b). What I did is as follows:

For $n = 2$, \begin{equation} a_{n+1} = a_{3} = \sqrt{a_{2} + a_{1}} = \sqrt{1+1} = \sqrt{2} > a_{2} = 1; \tag{1} \end{equation} For $n = k$, assume $a_{k+1} > a_{k}$;

For $n = k+1$, \begin{equation} a_{n+1} = a_{k+2} = \sqrt{a_{k+1} + a_{k}} > \sqrt{a_{k} + a_{k}} = \sqrt{2a_{k}} . \tag{2} \end{equation}

I need to prove $a_{k+2} > a_{k+1}$, but (2) does not lead to it. How should I solve this problem?

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The inductive hypothesis says $\;a_{k+1}\ge a_k\ge a_{k-1}\;$ , so

$$a_{k+2}:=\sqrt{a_{k+1}+a_{k}}\ge\sqrt{a_k+a_{k-1}}=:a_{k+1}\implies Q.E.D.$$

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Note that\begin{align}a_{n+2}>a_{n+1}&\iff\sqrt{a_{n+1}+a_n}+\sqrt{a_n+a_{n-1}}\\&\iff a_{n+1}+a_n>a_n+a_{n-1}\\&\iff a_{n+1}>a_{n-1},\end{align}which is true, since $a_{n+1}>a_n>a_{n-1}$.

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