Given,
$$\sqrt{2}-c_{n+1}= \sqrt{2}+\frac{c_n+2}{c_n+1}$$
I converted the form in R.H.S ,
$$\sqrt{2}+\frac{c_n+2}{c_n+1}=\frac{(\sqrt{2}+c_{n})(1+\sqrt{2})}{1+c_{n}}$$
then did as following,
$$\mid\sqrt{2}-c_{n+1}\mid <\frac{\mid\sqrt{2}-c_{n}\mid}{1+\sqrt{2}}$$
$$\Rightarrow \mid\frac{(\sqrt{2}+c_{n})(1+\sqrt{2})}{1+c_{n}}\mid <\frac{\mid\sqrt{2}-c_{n}\mid}{1+\sqrt{2}}$$
$$\Rightarrow \frac{\mid\sqrt{2}+c_{n}\mid\mid1+\sqrt{2}\mid}{\mid\sqrt{2}-c_{n}\mid\mid1+c_{n}\mid}<\frac{1}{1+\sqrt{2}}$$
I tried a lot, but got stuck here .How it can be proved after this? Or any other method to prove !!
Generally speaking, the inequality does not hold. Below I have put some sample values for $c_n$, $c_{n+1}= - \frac{c_n+2}{c_n+1}$, the infimum and the supremum of the set of values for $c_{n+1}$ where the inequality holds, and whether the inequality holds.
\begin{array}{cccccc} c_n & c_{n+1} & \text{Numerical}\ c_{n+1} & \inf & \sup & \text{Inequality} \\ -4 & -\frac{2}{3} & -0.666667 & 2-2 \sqrt{2} & 4 \sqrt{2}-2 & \text{True} \\ -3 & -\frac{1}{2} & -0.5 & 1-\sqrt{2} & 3 \sqrt{2}-1 & \text{False} \\ -\frac{5}{2} & -\frac{1}{3} & -0.333333 & \frac{1}{2}-\frac{1}{\sqrt{2}} & \frac{5}{\sqrt{2}}-\frac{1}{2} & \text{False} \\ -2 & 0 & 0. & 0 & 2 \sqrt{2} & \text{False} \\ -\frac{9}{5} & \frac{1}{4} & 0.25 & \frac{1}{5} \left(\sqrt{2}-1\right) & \frac{1}{5} \left(1+9 \sqrt{2}\right) & \text{True} \\ -\frac{3}{2} & 1 & 1. & \frac{1}{\sqrt{2}}-\frac{1}{2} & \frac{1}{2}+\frac{3}{\sqrt{2}} & \text{True} \\ -\frac{1}{2} & -3 & -3. & \frac{3}{2} \left(\sqrt{2}-1\right) & \frac{3}{2}+\frac{1}{\sqrt{2}} & \text{False} \\ 0 & -2 & -2. & 2 \left(\sqrt{2}-1\right) & 2 & \text{False} \\ 1 & -\frac{3}{2} & -1.5 & 3 \left(\sqrt{2}-1\right) & 3-\sqrt{2} & \text{False} \\ \sqrt{2} & -\sqrt{2} & -1.41421 & \sqrt{2} & \sqrt{2} & \text{False} \\ 2 & -\frac{4}{3} & -1.33333 & 4-2 \sqrt{2} & 4 \left(\sqrt{2}-1\right) & \text{False} \\ 3 & -\frac{5}{4} & -1.25 & 5-3 \sqrt{2} & 5 \left(\sqrt{2}-1\right) & \text{False} \\ \end{array}