Prove the inequality

[Math Tutor]

Proove that:
[tex]\frac{a_1}{(1+a_1)^2}+\frac{a_2}{(1+a_1+a_2)^2}+...+\frac{a_n}{(1+a_1+a_2+...+a_n)^2}\le \frac{a_1+a_1+...+a_n}{1+a_1+a_1+...+a_n}[/tex]

The numbers [tex]a_1, a_2, ... , a_n[/tex] are real and positive.

[Math Tutor]

Author: Knowledge Greedy

[Guest]

Spoiler: show

Follows from induction. It is trivial to check it is true for [tex]n=1[/tex]. Assuming it is true for [tex]n=k[/tex], add the trivially true inequality
[tex]\frac{a_{k+1}}{(1+a_1+\cdots+a_{k+1})^2}\leq \frac{a_{k+1}}{(1+a_1+\cdots+a_{k+1})(1+a_1+\cdots+a_k)}[/tex]
to the [tex]n=k[/tex] version of the inequality, simplifying should yield the case [tex]n=k+1[/tex]. (To make the simplification more obvious try using the substitution [tex]b=a_1+\cdots+a_k[/tex].)

R. Baber.