Question

Let $x_1, x_2, x_3, ....., x_{2018}$ be positive real numbers satisfying the condition

$\frac{1}{1+x_1} + \frac{1}{1+x_2} +....+ \frac{1}{1+x_{2018}} = 1$. Then the minimum value of $\prod_{r=1}^{2018} x_r$ is equals to $(k)^{k+1}$, where k is equal to

Answer 1

2017

Answer 2

Let $n = 2018$. The given condition is $\sum_{r=1}^{n} \frac{1}{1+x_r} = 1$, where $x_r > 0$.

Let $y_r = \frac{1}{1+x_r}$.

Since $x_r > 0$, we have $1+x_r > 1$, which implies $0 < y_r < 1$.

The given condition becomes $\sum_{r=1}^{n} y_r = 1$.

We need to find the minimum value of the product $\prod_{r=1}^{n} x_r$.

First, express $x_r$ in terms of $y_r$:

$y_r = \frac{1}{1+x_r} \implies 1+x_r = \frac{1}{y_r} \implies x_r = \frac{1}{y_r} - 1 = \frac{1-y_r}{y_r}$.

Now, substitute this into the product:

$P = \prod_{r=1}^{n} x_r = \prod_{r=1}^{n} \frac{1-y_r}{y_r}$.

Since $\sum_{i=1}^{n} y_i = 1$, we can write $1-y_r = \left(\sum_{i=1}^{n} y_i\right) - y_r = \sum_{i \neq r} y_i$.

So, the product becomes:

$P = \prod_{r=1}^{n} \frac{\sum_{i \neq r} y_i}{y_r} = \left(\frac{y_2+y_3+...+y_n}{y_1}\right) \left(\frac{y_1+y_3+...+y_n}{y_2}\right) ... \left(\frac{y_1+y_2+...+y_{n-1}}{y_n}\right)$.

Now, we apply the AM-GM inequality to the sum in each numerator. For any $r$, the sum $\sum_{i \neq r} y_i$ consists of $n-1$ positive terms.

By AM-GM inequality:

$\sum_{i \neq r} y_i \ge (n-1) \left(\prod_{i \neq r} y_i\right)^{\frac{1}{n-1}}$.

Applying this to each term in the product $P$:

$P \ge \prod_{r=1}^{n} \frac{(n-1) \left(\prod_{i \neq r} y_i\right)^{\frac{1}{n-1}}}{y_r}$

$P \ge (n-1)^n \prod_{r=1}^{n} \frac{\left(\prod_{i \neq r} y_i\right)^{\frac{1}{n-1}}}{y_r}$.

Let's analyze the product term:

$\prod_{r=1}^{n} \frac{\left(\prod_{i \neq r} y_i\right)^{\frac{1}{n-1}}}{y_r} = \frac{\left(\prod_{i \neq 1} y_i\right)^{\frac{1}{n-1}}}{y_1} \cdot \frac{\left(\prod_{i \neq 2} y_i\right)^{\frac{1}{n-1}}}{y_2} \cdot ... \cdot \frac{\left(\prod_{i \neq n} y_i\right)^{\frac{1}{n-1}}}{y_n}$.

Let $Y = \prod_{i=1}^{n} y_i$. Then $\prod_{i \neq r} y_i = \frac{Y}{y_r}$.

The expression becomes:

$\frac{\left(\frac{Y}{y_1}\right)^{\frac{1}{n-1}}}{y_1} \cdot \frac{\left(\frac{Y}{y_2}\right)^{\frac{1}{n-1}}}{y_2} \cdot ... \cdot \frac{\left(\frac{Y}{y_n}\right)^{\frac{1}{n-1}}}{y_n}$

$= \frac{Y^{\frac{1}{n-1}} y_1^{-\frac{1}{n-1}}}{y_1} \cdot \frac{Y^{\frac{1}{n-1}} y_2^{-\frac{1}{n-1}}}{y_2} \cdot ... \cdot \frac{Y^{\frac{1}{n-1}} y_n^{-\frac{1}{n-1}}}{y_n}$

$= Y^{\frac{n}{n-1}} \cdot (y_1 y_2 ... y_n)^{-\frac{1}{n-1}} \cdot (y_1 y_2 ... y_n)^{-1}$

$= Y^{\frac{n}{n-1}} \cdot Y^{-\frac{1}{n-1}} \cdot Y^{-1}$

$= Y^{\frac{n}{n-1} - \frac{1}{n-1} - 1} = Y^{\frac{n-1}{n-1} - 1} = Y^{1-1} = Y^0 = 1$.

So, the minimum value of $P$ is $(n-1)^n$.

The equality in AM-GM holds when all terms are equal. This means for each $r$, $y_i$ for $i \neq r$ must be equal. This implies $y_1=y_2=...=y_n$.

Given $\sum_{r=1}^{n} y_r = 1$, if $y_1=y_2=...=y_n$, then $n y_r = 1$, so $y_r = \frac{1}{n}$ for all $r$.

When $y_r = \frac{1}{n}$:

$x_r = \frac{1-y_r}{y_r} = \frac{1 - \frac{1}{n}}{\frac{1}{n}} = \frac{\frac{n-1}{n}}{\frac{1}{n}} = n-1$.

So, $x_r = n-1$ for all $r$.

The product $\prod_{r=1}^{n} x_r = (n-1)^n$.

Given $n=2018$.

The minimum value is $(2018-1)^{2018} = (2017)^{2018}$.

The question states that the minimum value is equal to $(k)^{k+1}$.

Comparing $(2017)^{2018}$ with $(k)^{k+1}$:

We can see that $k=2017$.

Then $k+1 = 2017+1 = 2018$.

So, $(k)^{k+1} = (2017)^{2018}$.

Thus, $k=2017$.