Given the real numbers $a_1, a_2,...,a_n$ all greater than $1$, such that $\prod_{i=1}^{n} a_i=10^n$, prove that: $$\frac{lga_1}{(1+lga_1)^2}+\frac{lga_2}{(1+lga_1a_2)^2}+...+\frac{lga_n}{(1+\sum_{i=1}^{n} lga_i)^2}\le\frac{n}{n+1}$$

My first try was with proof by induction. It did not work. My second attempt goes as follows: Let`s define the numbers $x_1,x_2,...,x_n$, such that $x_1=lga_1, x_2=lga_2,...,x_n=lga_n$. This implies that every number $x_i>0$, where $i$ is a random natural number; and that $\sum_{i=1}^{n} x_i=n$, so the inequality becomes: $$\frac{x_1}{(1+x_1)^2}+\frac{x_2}{(1+x_1 +x_2)^2}+...+\frac{x_n}{(1+\sum_{i=1}^{n} x_i)^2}=\frac{x_1}{(1+x_1)^2}+\frac{x_2}{(1+x_1+ x_2)^2}+...+\frac{x_n}{(1+n)^2}\le\frac{n}{n+1}$$ Then I tried to identify another expression for each of the fractions, so that we can work with them. My best guess was that for every $x_1,x_2,...,x_n$, $$\frac{x_n}{(1+\sum_{i=1}^{n} x_n)^2}< \frac{1}{1+\sum_{i=1}^{n} x_n}$$ so that the inequality can be rewritten as: $$\frac{1}{1+x_1}+\frac{1}{1+x_1+x_2}+...+\frac{1}{1+\sum_{i=1}^{n} x_n}\le\frac{n}{n+1}$$ But now I am stuck. My biggest problem is that the denominators contain multiple values of $x_n$. Is there a way or strategy to handle such problems more efficiently?

  • $\begingroup$ Is it provided that $a_i>a_j$ if $i>j$? $\endgroup$
    – Gwen
    May 15 at 14:26
  • $\begingroup$ No. It is not provided. $\endgroup$
    – fikooo
    May 15 at 14:29
  • $\begingroup$ Also by $\operatorname{lg}$ do you mean logarithm for base 10? $\endgroup$
    – Gwen
    May 15 at 14:30
  • $\begingroup$ Yes, logarithm for base 10, and I also edited the problem...I misinterpreted the problem, sorry $\endgroup$
    – fikooo
    May 15 at 14:32

2 Answers 2


The inequality is still true if the condition $x_1 + x_2 + \cdots + x_n = n$ is dropped. In other words, we have the following result.

Fact 1. Let $x_1, x_2, \cdots, x_n \ge 0$. Then $$\frac{x_1}{(1+x_1)^2}+\frac{x_2}{(1+x_1 +x_2)^2} + \cdots + \frac{x_n}{(1 + x_1 + x_2 + \cdots + x_n)^2} \le \frac{n}{1+n}.$$

Proof of Fact 1.

We use the Mathematical Induction.

When $n = 1$, true.

Assume that the statement is true for $n$ ($n\ge 1$).

For $n + 1$, we have \begin{align*} &\frac{x_1}{(1+x_1)^2}+\frac{x_2}{(1+x_1 +x_2)^2} + \cdots + \frac{x_{n+1}}{(1 + x_1 + x_2 + \cdots + x_{n+1})^2}\\ ={}& \frac{x_1}{(1+x_1)^2} + \frac{1}{1 + x_1}\\ &\quad \times \left(\frac{y_1}{(1+y_1)^2}+\frac{y_2}{(1+y_1 +y_2)^2} + \cdots + \frac{y_n}{(1 + y_1 + y_2 + \cdots + y_n)^2} \right)\\ \le{}& \frac{x_1}{(1+x_1)^2} + \frac{1}{1 + x_1}\cdot\frac{n}{n+1}\\ \le{}&\frac{(2n+1)^2}{4(n+1)^2}\\ \le{}& \frac{n+1}{n+2} \end{align*} where $y_k = \frac{x_{k+1}}{1 + x_1}, k=1, 2, \cdots, n$, and we use $$\frac{(2n+1)^2}{4(n+1)^2} - \frac{x_1}{(1+x_1)^2} - \frac{n}{n+1}\cdot \frac{1}{1 + x_1} = \frac{(2nx_1 + x_1 - 1)^2}{4(n+1)^2(1+x_1)^2}.$$

Thus, the statement is true for $n+1$.

We are done.

  • $\begingroup$ I don't quite get this. Shouldn't the inequality via induction look like $\frac{n}{n+1}+\frac{x_k}{(1+n+x_k)^2}\le\frac{n+1}{n+2}$, where $k=n+1$? $\endgroup$
    – fikooo
    May 19 at 10:52
  • $\begingroup$ @fikooo I discard the condition $x_1 + x_2 + \cdots + x_n = n$. $\endgroup$
    – River Li
    May 19 at 11:34
  • $\begingroup$ Would it work the same way using the condition? $\endgroup$
    – fikooo
    May 19 at 11:40
  • $\begingroup$ @fikooo How do you use the condition in Induction? $\endgroup$
    – River Li
    May 19 at 11:45
  • $\begingroup$ @fikooo You can take $n = 4$ to see the Mathematical Induction process. $\endgroup$
    – River Li
    May 19 at 12:20

The inequality on the last line of your post is false while the original inequality is true. The incorrect inequality sign should reverse. It is easy to see it:

$\dfrac{1}{1+x_1} \ge \dfrac{1}{1+x_1+x_2+....+x_n} = \dfrac{1}{1+n}$. Repeat this for the remaining terms, and add them up ! we have:

$\displaystyle \sum_{k=1}^n \dfrac{1}{1+\displaystyle \sum_{i=1}^k x_i}\ge\displaystyle \sum_{k=1}^n \dfrac{1}{1+n}=\dfrac{n}{1+n}. $

  • $\begingroup$ I was thinking about this solution as well, but I missed something...but I don't get it...which inequality's sign should be reversed? $\endgroup$
    – fikooo
    May 16 at 4:08
  • $\begingroup$ @fikooo: The one on the last line of your post. But the original one is valid. I have seen it somewhere. $\endgroup$
    – Wang YeFei
    May 16 at 17:06

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