We have $$S_{n+1}=S_n+\frac{4}{4+S_n}\implies a_{n+1}=\frac{4}{4+S_n}=\frac{4}{4+a_1+a_2+...+a_n}$$for all $n\in \mathbb{Z}^+$. Let $P(n)$ denote the statement $$a_{n}\geqslant \frac{4}{\sqrt{9n+7}}\text{ and }\sqrt{9n+16}-4\geqslant a_1+a_2+...+a_n.$$Since $a_1=1,a_2=\frac{4}{5},a_3=\frac{20}{29}$, $P(1),P(2),P(3)$ are true. For $m\geqslant 3$ suppose that $P(m)$ is true, we will show that $P(m+1)$ is true. Since $4+a_1+a_2+...+a_m\leqslant \sqrt{9m+16}$, $a_{m+1}\geqslant \frac{4}{\sqrt{9m+16}}=\frac{4}{\sqrt{9(m+1)+7}}.$ Then we need to prove that $\sqrt{9m+25}-4 \geqslant \sum_{i=1}^{m+1}{a_i} = t+\frac{4}{4+t}$ where $t=\sum_{i=1}^{m}{a_i} \leqslant \sqrt{9m+16}-4$. Note that $t\geqslant a_1+a_2+a_3>2$. Let $f(x)=x+\frac{4}{4+x}$. We get $f'(x)=\frac{x^2+8x-16}{(x+4)^2}>0$ when $x>2$. So, when $2< t\leqslant \sqrt{9m+16}-4$, we have $$f(t) = t+\frac{4}{4+t}\leqslant f( \sqrt{9m+16}-4)= \sqrt{9m+16}-4+\frac{4}{\sqrt{9m+16}}.$$We need $$ \sqrt{9m+16}-4+\frac{4}{\sqrt{9m+16}} \leqslant \sqrt{9m+25}-4,$$which is equivalent to $$\sqrt{9m+25} \geqslant \frac{9m+20}{\sqrt{9m+16}} \iff (9m+25)(9m+16)\geqslant (9m+20)^2,$$which is true. This completes the induction step and we are done.

Solution,

Vietnamisalwaysinmyheart wrote: Solution,

Beautiful.

We can easily find that $a_{n+1}=\frac{4}{4+S_n}$ , so it's enough to show that $S_n\leq \sqrt{9m+16}-4$ , and it's easy to proof with induction