# An inequality involving positive real numbers

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

$$\frac{a_1}{(a_1+\cdots+a_n)\log^2(a_1+\cdots+a_{n}+1)}+\frac{a_2}{(a_2+\cdots+a_n)\log^2(a_2+\cdots+a_n+1)}+\cdots+\frac{a_n}{a_n\log^2(a_n+1)} \le C$$ 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 $$C=4$$.

• Can you add to your question some examples you tried, and/or the source of the question? It might help to have more context. Aug 29, 2020 at 1:42
• The usual trickery: consider the function $f(x)=\frac{1}{x\log^2 (1+x)}$ and the points $b_j=a_n+a_{n-1}+\dots+a_{n-j}$. Then the $j+1$-th term from the end is at most $\int_{b_{j-1}}^{b_j}f(x)\,dx$, so the whole sum is at most $\frac{1}{\log^2 2}+\int_1^\infty f(x)\,dx$. Aug 29, 2020 at 4:14
• @fedja Would you mind writing that as an answer? Aug 29, 2020 at 6:28
• @fedia: You solution is correct. Can you put it as an answer? Otherwise the question is listed as "unanswered". Aug 30, 2020 at 1:49
• @JCAA "Can you lower it to 4?" No, the constant is sharp. Take $a_n=1$ and go sufficiently far by very small steps afterwards. Then the Riemann sum will be as close to the integral as you wish. Aug 30, 2020 at 5:35

The usual trickery: consider the function $$f(x)=\frac1{x\log^2(x+1)}$$ 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 $$\frac 1{\log^2 2}+\int_1^\infty f(x) dx\le 4.075$$.