navi_09220114 wrote: Given $n$ positive real numbers $x_1,x_2,x_3,...,x_n$ such that $$\left (1+\frac{1}{x_1}\right )\left(1+\frac{1}{x_2}\right)...\left(1+\frac{1}{x_n}\right)=(n+1)^n$$Determine the minimum value of $x_1+x_2+x_3+...+x_n$. this it is beautiful! Hint let $f(x)=\ln(1+\frac{1}{x})$ and Jensen inequality.

A slightly different solution based on the AM-GM-HM inequality. From the given statement, we have $$\frac{1}{n+1} = \frac{1}{(1+\frac{1}{x_1})^\frac{1}{n}(1+\frac{1}{x_2})^\frac{1}{n}\cdots(1+\frac{1}{x_2})^\frac{1}{n}}$$$$\leq \frac{1}{n} (\frac{1}{1+\frac{1}{x_1}}+\frac{1}{1+\frac{1}{x_2}}+\cdots+\frac{1}{1+\frac{1}{x_n}})$$$$ = \frac{1}{n} (\frac{x_1}{x_1+1}+\frac{x_2}{x_2+1}+\cdots+\frac{x_n}{x_n+1})$$$$ = \frac{1}{n}(n- (\frac{1}{x_1+1}+\frac{1}{x_2+1}+\cdots+\frac{1}{x_n+1})) $$where the second line follows by AM-GM inequality. From the inequality above, we then have $$ 1 - \frac{1}{n+1} \geq \frac{1}{n}(\frac{1}{x_1+1}+\frac{1}{x_2+1}+\cdots+\frac{1}{x_n+1}) \geq \frac{n}{x_1+x_2+\cdots+x_n+n}$$where the last inequality follows by AM-HM inequality. Simplifying yields $x_1+\cdots+x_n \geq1$, attained when all the $x_i$ are equal.

We start with the equation: \[ \prod_{i=1}^n \left(1 + \frac{1}{x_i}\right) = (n+1)^n. \] Taking the natural logarithm on both sides: \[ \ln \left( \prod_{i=1}^n \left(1 + \frac{1}{x_i}\right) \right) = \ln \left((n+1)^n\right). \] \[ \sum_{i=1}^n \ln\left(1 + \frac{1}{x_i}\right) = n \ln(n+1). \]=> \[ \sum_{i=1}^n \ln\left(1 + \frac{1}{x_i}\right) = \sum_{i=1}^n \ln(n+1). \]Let \( f(x) = \ln\left(1 + \frac{1}{x}\right) \). The second derivative of \( f(x) \) is: \[ f''(x) = \frac{2x + 1}{(x^2 + x)^2}. \] Since \( f''(x) > 0 \) for \( x > 0 \), \( f(x) \) is a convex function. Now, applying Jensen's inequality for a convex function, we have: \[ \ln\left(1 + \frac{1}{\frac{1}{n} \sum_{i=1}^n x_i}\right) \leq \frac{1}{n} \sum_{i=1}^n \ln\left(1 + \frac{1}{x_i}\right). \] From the given equation: \[ \frac{1}{n} \sum_{i=1}^n \ln\left(1 + \frac{1}{x_i}\right) = \ln(n+1). \] Substituting this back, we get: \[ \ln\left(1 + \frac{1}{\frac{1}{n} \sum_{i=1}^n x_i}\right) \leq \ln(n+1). \] Exponentiating both sides: \[ 1 + \frac{1}{\frac{1}{n} \sum_{i=1}^n x_i} \leq n+1. \] Simplify: \[ \frac{1}{\frac{1}{n} \sum_{i=1}^n x_i} \leq n. \] Taking the reciprocal: \[ \frac{1}{n} \sum_{i=1}^n x_i \geq \frac{1}{n}. \] Multiplying through by \( n \), we find: \[ \sum_{i=1}^n x_i \geq 1. \] Equality holds in Jensen's inequality if and only if \( x_i = \frac{1}{n} \) for all \( i \in \{1, 2, \ldots, n\} \).

kznil96 wrote: A slightly different solution based on the AM-GM-HM inequality. From the given statement, we have $$\frac{1}{n+1} = \frac{1}{(1+\frac{1}{x_1})^\frac{1}{n}(1+\frac{1}{x_2})^\frac{1}{n}\cdots(1+\frac{1}{x_2})^\frac{1}{n}}$$$$\leq \frac{1}{n} (\frac{1}{1+\frac{1}{x_1}}+\frac{1}{1+\frac{1}{x_2}}+\cdots+\frac{1}{1+\frac{1}{x_n}})$$$$ = \frac{1}{n} (\frac{x_1}{x_1+1}+\frac{x_2}{x_2+1}+\cdots+\frac{x_n}{x_n+1})$$$$ = \frac{1}{n}(n- (\frac{1}{x_1+1}+\frac{1}{x_2+1}+\cdots+\frac{1}{x_n+1})) $$where the second line follows by AM-GM inequality. From the inequality above, we then have $$ 1 - \frac{1}{n+1} \geq \frac{1}{n}(\frac{1}{x_1+1}+\frac{1}{x_2+1}+\cdots+\frac{1}{x_n+1}) \geq \frac{n}{x_1+x_2+\cdots+x_n+n}$$where the last inequality follows by AM-HM inequality. Simplifying yields $x_1+\cdots+x_n \geq1$, attained when all the $x_i$ are equal. Yes! that it is elaborate solution!

Assume \( x_1 + \cdots + x_n < 1 \), then by AM-GM \[ \prod \left(1 + \frac{1}{x_i}\right) > \prod \left(1 + \frac{x_1 + \cdots + x_n}{x_i}\right) = \prod \frac{x_i + x_1 + \cdots + x_n}{x_i} \geq \prod \frac{(n+1) \sqrt[n+1]{x_i \cdot x_1 \cdots x_n}}{x_i} = (n+1)^n. \] Contradiction, this indicates that \(x_1 + \cdots + x_n \geq 1\), with the equality holding when all \(x_i\) are equal.

lbh_qys wrote: Assume \( x_1 + \cdots + x_n < 1 \), then by AM-GM \[ \prod \left(1 + \frac{1}{x_i}\right) > \prod \left(1 + \frac{x_1 + \cdots + x_n}{x_i}\right) = \prod \frac{x_i + x_1 + \cdots + x_n}{x_i} \geq \prod \frac{(n+1) \sqrt[n+1]{x_i \cdot x_1 \cdots x_n}}{x_i} = (n+1)^n. \] Contradiction, this indicates that \(x_1 + \cdots + x_n \geq 1\), with the equality holding when all \(x_i\) are equal. wow! solution!

lbh_qys wrote: Assume \( x_1 + \cdots + x_n < 1 \), then by AM-GM \[ \prod \left(1 + \frac{1}{x_i}\right) > \prod \left(1 + \frac{x_1 + \cdots + x_n}{x_i}\right) = \prod \frac{x_i + x_1 + \cdots + x_n}{x_i} \geq \prod \frac{(n+1) \sqrt[n+1]{x_i \cdot x_1 \cdots x_n}}{x_i} = (n+1)^n. \] Contradiction, this indicates that \(x_1 + \cdots + x_n \geq 1\), with the equality holding when all \(x_i\) are equal. nice!!