Positive sequence $\{a_n\}$, $a_{n+1}=-\sqrt{a_n}+1, a_1\in{(\frac12,1)}$. Prove $a_{2n+1}+a_{2n+2}<a_{2n-1}+a_{2n}$.

Question

Positive sequence $\{a_n\}$, $a_{n+1}=-\sqrt{a_n}+1, a_1\in{(\frac12,1)}$. Prove

$a_{2n+1}+a_{2n+2}<a_{2n-1}+a_{2n}$.

I have proved that $\{a_{2n+1}\}$ is decreasing while $\{a_{2n}\}$ is increasing and they both converge to $(\frac{\sqrt 5-1}{2})^2$. I tried to prove the inequality by induction but it seems hard. Could anyone help me out? Thanks.

Answer 1

By simple evaluation we have $$ \begin{align} a_{2n+2}-a_{2n} &= (1-\sqrt{a_{2n+1}})-(1-\sqrt{a_{2n-1}})\\ &= \sqrt{a_{2n-1}}-\sqrt{a_{2n+1}} \end{align}$$ Now we will prove that $\sqrt{a_{2n-1}}-\sqrt{a_{2n+1}} \le a_{2n-1}-a_{2n+1}$, which is equivalent to $\sqrt{a_{2n-1}}-a_{2n-1}\le \sqrt{a_{2n+1}}-a_{2n+1}$. Now apply following facts:

1. $(\frac{\sqrt 5-1}{2})^2 < a_{2n-1}<1$,
2. $f(x)=\sqrt{x}-x$ is decreasing over $[1/4,1]$ and
3. $1/4 < (\frac{\sqrt 5-1}{2})^2$.