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

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