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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 $C=4$.

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    $\begingroup$ Can you add to your question some examples you tried, and/or the source of the question? It might help to have more context. $\endgroup$ Aug 29, 2020 at 1:42
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    $\begingroup$ 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 $\frac{1}{\log^2 2}+\int_1^\infty f(x)\,dx$. $\endgroup$
    – fedja
    Aug 29, 2020 at 4:14
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    $\begingroup$ @fedja Would you mind writing that as an answer? $\endgroup$ Aug 29, 2020 at 6:28
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    $\begingroup$ @fedia: You solution is correct. Can you put it as an answer? Otherwise the question is listed as "unanswered". $\endgroup$
    – markvs
    Aug 30, 2020 at 1:49
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    $\begingroup$ @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. $\endgroup$
    – fedja
    Aug 30, 2020 at 5:35

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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 $f(x)=\frac1{x\log^2(x+1)}$ 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 $\frac 1{\log^2 2}+\int_1^\infty f(x) dx\le 4.075$.

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