If $a_1a_2a_3...a_{20}=1$ find minimum value of S Given that
If $$a_1a_2a_3...a_{20}=1$$
then what is the minimum value of 
$$\frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+...+\frac{a_{20}}{(1+a_1)(1+a_2)...(1+a_{20})}=S$$
Please Suggest me how to start?
 A: Define a more general sum:
$$S_n=\frac{a_1}{1+a_1}+\frac{a_2}{(1+a_1)(1+a_2)}+...+\frac{a_{n}}{(1+a_1)(1+a_2)...(1+a_{n})}$$
In your particular case $n=20$. Let us first show that:
$$S_n=1-\frac{1}{\prod_{i=1}^n(1+a_i)}\tag{1}$$
You can guess that formula by adding a few starting items and you can rigorously prove it by induction. It is obviously true for $n=1$:
$$S_1=1-\frac{1}{1+a_1}=\frac{a_1}{1+a_1}$$
On the other side:
$$S_{n+1}=S_n+\frac{a_{n+1}}{(1+a_1)(1+a_2)...(1+a_{n})(1+a_{n+1})}$$
$$S_{n+1}=1-\frac{1}{\prod_{i=1}^n(1+a_i)}+\frac{a_{n+1}}{\prod_{i=1}^{n+1}(1+a_i)}$$
$$S_{n+1}=1-\frac{1+a_{n+1}}{\prod_{i=1}^{n+1}(1+a_i)}+\frac{a_{n+1}}{\prod_{i=1}^{n+1}(1+a_i)}$$
$$S_{n+1}=1-\frac{1}{\prod_{i=1}^{n+1}(1+a_i)}$$
So (1) is definitely correct. 
To minimize $S_n$ you have to minimize the product:
$$P_n=\prod_{i=1}^{n}(1+a_i)$$
By simple AM-GM inequality we know that:
$$1+a_i\ge2\sqrt{a_i}$$
Equality is reached only for $a_i=1$. It follows that:
$$P_n=\prod_{i=1}^{n}(1+a_i)\ge2^n\sqrt{\prod_{i=1}^{n}a_i}$$
Assuming that $\prod_{i=1}^{n}a_i=1$:
$$P_n\ge2^n$$
$$S_n\ge1-\frac{1}{P_n}$$
$$S_n\ge1-\frac{1}{2^n}$$
Minimum value:
$$(S_n)_{min}=1-\frac{1}{2^n}$$
...is reached for $a_1=a_2=\dots=a_n=1$
