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

Question

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: $$\frac{\log_{10}a_1}{(1+\log_{10}a_1)^2}+\frac{\log_{10}a_2}{(1+\log_{10}a_1 + \log_{10}a_2)^2}+...+\frac{\log_{10}a_n}{(1+\sum_{i=1}^{n} \log_{10}a_i)^2}\le\frac{n}{n+1}$$

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 $x_1=\log_{10}a_1, x_2=\log_{10}a_2,...,x_n=\log_{10}a_n$. This implies that every number $x_i>0$, where $i=1,2,\cdots,n$ ; and that $\sum_{i=1}^{n} x_i=n$, so the inequality becomes: $$\frac{x_1}{(1+x_1)^2}+\frac{x_2}{(1+x_1 +x_2)^2}+...+\frac{x_n}{(1+\sum_{i=1}^{n} x_i)^2}\le\frac{n}{n+1}.$$ $\phantom{2}$

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?

Answer 1

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.

Answer 2

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}. $