Prove the following bound for the following sequence Define a sequence by $a_1= \frac13$ and $a_{n+1}=a_n(1-a_n)$ for $n=1,2,\dots$.
How can I show that $a_n\geq \frac{1}{3n}$?
I proceed by induction: 
Im stuck in the following:
We have that $0\leq a_n \leq \frac 13$. Then I have that $a_{n+1}\geq \frac 29$. But I need to get $a_{n+1}\geq \frac{1}{3(n+1)}$.
Any ideas or another way to do it, I will appreciate? 
 A: I may be missing a much easier proof here, but I believe you can prove 
$$\frac{1-\sqrt{(3n-1)/(3n+3)}}{2} \leq a_n \leq \frac{1+\sqrt{(3n-1)/(3n+3)}}{2}.\;\;\; (*)$$
The trick here is to realize that if the above is true, then it is true that
$$\left(a_n - \frac{1-\sqrt{(3n-1)/(3n+3)}}{2} \right)\left(a_n - \frac{1+\sqrt{(3n-1)/(3n+3)}}{2} \right) \leq 0.$$
Upon rearranging, this is equivalent to
$$a_n(1-a_n) \geq \frac{1}{3(n+1)}.$$
That is, proving $(*)$ for $a_n$ implies that $a_{n+1}\geq \frac{1}{3(n+1)}$. We use induction to prove $(*)$.
The base case $n=1$ is true. In the inductive step, you need to prove the estimate
$$a_{n+1}\leq\left(\frac{1+\sqrt{(3n-1)/(3n+3)}}{2}\right)\left(1-\frac{1}{3n}\right)\leq \frac{1+\sqrt{(3n+2)/(3n+6)}}{2}.$$
Note, we also use here that $a_n\geq \frac{1}{3n}$, which is true by the inductive hypothesis $(*)$ on $a_{n-1}$.
This inequality follows from the fact that the function 
$$f(x) := \frac{1+\sqrt{(3x+2)/(3x+6)}}{2}-\left(\frac{1+\sqrt{(3x-1)/(3x+3)}}{2}\right)\left(1-\frac{1}{3x}\right)$$
is monotone, satisfies $f(1)>0$ and $\lim_{x\to \infty} f(x) = 0$. 
To conclude we establish the estimate
$$a_{n+1} \geq \left(\frac{1-\sqrt{(3n-1)/(3n+3)}}{2}\right)\left(\frac{1-\sqrt{(3n-1)/(3n+3)}}{2}\right) \geq \left(\frac{1-\sqrt{(3n+2)/(3n+6)}}{2}\right).$$
Again, here we show the second inequality by constructing an associated function (as above) and showing that it is monotone decreasing to 0.
I admit that I have only checked these associated functions graphically, but I believe this proof method should be sufficient. I would be very interested in a more elementary proof.
