Mahler's Inequality

From ProofWiki
Jump to: navigation, search

Theorem

The geometric mean of the termwise sum of two finite sequences of positive numbers is never less than the sum of their two separate geometric means:

$\displaystyle \prod_{k=1}^n \left({x_k + y_k}\right)^{1/n} \ge \prod_{k=1}^n x_k^{1/n} + \prod_{k=1}^n y_k^{1/n}$

where $x_k, y_k > 0$ for all $k$.


Proof

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \prod_{k=1}^n \left({\frac {x_k} {x_k + y_k} }\right)^{1/n}\) \(\le\) \(\displaystyle \frac 1 n \sum_{k=1}^n \frac {x_k} {x_k + y_k}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Arithmetic Mean Never Less than Geometric Mean          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \prod_{k=1}^n \left({\frac {y_k} {x_k + y_k} }\right)^{1/n}\) \(\le\) \(\displaystyle \frac 1 n \sum_{k=1}^n \frac {y_k} {x_k + y_k}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          Arithmetic Mean Never Less than Geometric Mean          
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle \prod_{k=1}^n \left({\frac {x_k} {x_k + y_k} }\right)^{1/n} + \prod_{k=1}^n \left({\frac {y_k} {x_k + y_k} }\right)^{1/n}\) \(\le\) \(\displaystyle \frac 1 n \sum_{k=1}^n \frac {x_k} {x_k + y_k} + \frac 1 n \sum_{k=1}^n \frac {y_k} {x_k + y_k}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)          adding them together          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \frac 1 n \sum_{k=1}^n \frac {x_k + y_k} {x_k + y_k}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \frac 1 n n\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle 1\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    

This leads to:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \prod_{k=1}^n \left({\frac {x_k} {x_k + y_k} }\right)^{1/n} + \prod_{k=1}^n \left({\frac {y_k} {x_k + y_k} }\right)^{1/n}\) \(\le\) \(\displaystyle 1\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle \frac {\displaystyle \prod_{k=1}^n \left({x_k}\right)^{1/n} } {\displaystyle \prod_{k=1}^n \left({x_k + y_k}\right)^{1/n} } + \frac {\displaystyle \prod_{k=1}^n \left({y_k}\right)^{1/n} } {\displaystyle \prod_{k=1}^n \left({x_k + y_k}\right)^{1/n} }\) \(\le\) \(\displaystyle 1\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle \frac {\displaystyle \prod_{k=1}^n \left({x_k}\right)^{1/n} + \prod_{k=1}^n \left({y_k}\right)^{1/n} } {\displaystyle \prod_{k=1}^n \left({x_k + y_k}\right)^{1/n} }\) \(\le\) \(\displaystyle 1\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle \prod_{k=1}^n \left({x_k}\right)^{1/n} + \prod_{k=1}^n \left({y_k}\right)^{1/n}\) \(\le\) \(\displaystyle \prod_{k=1}^n \left({x_k + y_k}\right)^{1/n}\) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \)                    

$\blacksquare$


Source of Name

This entry was named for Kurt Mahler.

Personal tools
Namespaces
Variants
Actions
Navigation
ProofWiki.org
ToDo
Toolbox
Google AdSense