<s>[INST] For $ f_{n+1} = \sin f_n(x) $, $f_n(x) \gt f_2(x)/(n-1) $? 
Let $f_0(x) = x, f_{n+1} = \sin f_n(x) (0 < x < \pi)$.
  Then for $n \ge 3$,
  $$f_n(x) \gt \frac{f_2(x)}{n-1} ? $$

Here is what I have shown:
For $x > 0$, $\sin x > x - x^3 / 6$.
So to show the proposition, it suffices to show that, by induction, for $n$, $n f_n(x)^2 < 6$.
And so it suffices to show that $n f_n(1)^2 < 6$.
And using a calculator, this inequality seems to be true.
 [/INST] Analogously to $\sin(x)<x-x^3/6$, we have $\sin(x)>x-x^3/6+x^5/120$ (one additional term in the Taylor Series). Additionally, $\sin([0, 1])\subseteq [0,1]$ so that $0\leq f_n(1)\leq 1$ for all $n$.
We prove by induction that $nf_n(1)^2<6$. Base case: $f_1(1)^2=\sin(1)<6$. Now let $n\geq 2$ and assume the statement holds for $n-1$. Since $\sin(x)$ is increasing on $[0,1]$, we have
$$nf_n(1)^2=n\sin^2(f_{n-1}(1))<n\sin^2\left(\sqrt{\frac{6}{n-1}}\right)<n\left(\sqrt{\frac{6}{n-1}}-\frac{1}{n-1}\sqrt{\frac{6}{n-1}}+\frac{3}{10(n-1)^2}\sqrt{\frac{6}{n-1}}\right)^2=6\left(\sqrt{\frac{n}{n-1}}-\frac{1}{n-1}\sqrt{\frac{n}{n-1}}+\frac{3}{10(n-1)^2}\sqrt{\frac{n}{n-1}}\right)=6g(n),$$
where
$$g(x)=\sqrt{\frac{x}{x-1}}-\frac{1}{x-1}\sqrt{\frac{x}{x-1}}+\frac{3}{10(x-1)^2}\sqrt{\frac{x}{x-1}}.$$
We have to prove $g(n)\leq 1$ for all $n\geq 2$.
But the only local extremum of $g(x)$ in $\mathbb{R}^+$ is between $x=1$ and $x=2$ and there $g(x)<1$. Furthermore, we easily see
$\lim\limits_{x\rightarrow\infty} g(x)=1,$
so that $g(x)\leq 1$ for all $x\geq 2$. Thus, the claim is proved.
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