# An inequality involving positive real numbers

Mathematics Asked by JCAA on November 14, 2020

I need to prove the following inequality. Let $$a_1, …, a_n$$ be positive real numbers with $$a_n ge 1$$. Then

$$frac{a_1}{(a_1+cdots+a_n)log^2(a_1+cdots+a_{n}+1)}+frac{a_2}{(a_2+cdots+a_n)log^2(a_2+cdots+a_n+1)}+cdots+frac{a_n}{a_nlog^2(a_n+1)} le C$$
for some constant $$C$$ which does not depend on $$a_i$$‘s. This seems to be true but non-trivial.

Edit The constant $$C$$ should not depend on $$n$$. I suspect that one can take $$C=4$$.

This answer is a copy of @fedja's comment above. I make it an answer to change the status of question to "have an accepted answer". If fedja wants he can post his own answer, then I will delete this answer and accept his.

The usual trickery: consider the function $$f(x)=frac1{xlog^2(x+1)}$$ and the points $$b_j=a_n+...+a_{n-j}$$. Then the $$j+1$$st term from the end is at most $$int_{b_{j-1}}^{b_j}f(x) dx$$, so the whole sum is at most $$frac 1{log^2 2}+int_1^infty f(x) dxle 4.075$$.

Correct answer by JCAA on November 14, 2020