There is a sequence S defined as follows: $S_1=\frac{1}{\sqrt3}$, $S_{n+1}=\frac{S_n}{1+\sqrt{1+S_n^2}}$ for all $n≥1.$
Let $t_n$ be the same recurrence as $S_n$ but with $t_1>\frac{1}{\sqrt{3}}$, prove that $t_n>S_n$ for all $n≥1.$
I have tried to use mathematical induction, but I cannot figure out how to show $t_{n+1}>S_{n+1}$. Any help or hints would be greatly appreciated.
To show that $$0\lt S_n\lt t_n\Rightarrow S_{n+1}\lt t_{n+1}$$
You need to show that $$\frac{S_n}{1+\sqrt{1+S_n^2}}\lt \frac{t_n}{1+\sqrt{1+t_n^2}}$$ $$S_n+S_n\sqrt{1+t_n^2}\lt t_n+t_n\sqrt{1+S_n^2}$$
Since $S_n\lt t_n$, then is enough to show that $$S_n\sqrt{1+t_n^2}\lt t_n\sqrt{1+S_n^2}|()^2$$
The inequality holds for positive numbers so square each side:
$$S_n^2+S_n^2t_n^2\lt t_n^2+S_n^2t_n^2$$ $$S_n^2\lt t_n^2$$ Which is true since $S_n\lt t_n$