# Logarithmic inequality involving $a_1, a_2, ..., a_n$

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?

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

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.

• 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 $k=n+1$? May 19 at 10:52
• @fikooo I discard the condition $x_1 + x_2 + \cdots + x_n = n$. May 19 at 11:34
• Would it work the same way using the condition? May 19 at 11:40
• @fikooo How do you use the condition in Induction? May 19 at 11:45
• @fikooo You can take $n = 4$ to see the Mathematical Induction process. 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:

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

• 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? May 16 at 4:08
• @fikooo: The one on the last line of your post. But the original one is valid. I have seen it somewhere. May 16 at 17:06