proof of arithmetic-geometric-harmonic means inequality

For the Arithmetic Geometric Inequality, I claim it is enough to prove that if $\prod_{i=1}^{n}x_{i}=1$ with $x_{i}\geq 0$ then $\sum_{i=1}^{n}x_{i}\geq n$ . The arithmetic geometric inequality for $y_{1},\ldots,y_{n}$ will follow by taking $x_{i}=\frac{y_{i}}{\sqrt[n]{\prod_{k=1}^{n}y_{k}}}$ . The geometric harmonic inequality follows from the arithmetic geometric by taking $x_{i}=\frac{1}{y_{i}}$ .

So, we show that if $\prod_{i=1}^{n}x_{i}=1$ with $x_{i}\geq 0$ then $\sum_{i=1}^{n}x_{i}\geq n$ by induction on $n$ .

Clear for $n=1$ .

Induction Step: By reordering indices we may assume the $x_{i}$ are increasing, so $x_{n}\geq 1\geq x_{1}$ . Assuming the statement is true for $n-1$ , we have $x_{2}+\cdots+x_{n-1}+x_{1}x_{n}\geq n-1$ . Then,

\sum_{i=1}^{n}x_{i}\geq n-1+x_{n}+x_{1}-x_{1}x_{n}

by adding $x_{1}+x_{n}$ to both sides and subtracting $x_{1}x_{n}$ . And so,

	$\displaystyle\sum_{i=1}^{n}x_{i}$	$\displaystyle\geq n+(x_{n}-1)+(x_{1}-x_{1}x_{n})$
		$\displaystyle=n+(x_{n}-1)-x_{1}(x_{n}-1)$
		$\displaystyle=n+(x_{n}-1)(1-x_{1})$
		$\displaystyle\geq n$

The last line follows since $x_{n}\geq 1\geq x_{1}$ .

Title	proof of arithmetic-geometric-harmonic means inequality
Canonical name	ProofOfArithmeticgeometricharmonicMeansInequality
Date of creation	2013-03-22 15:09:37
Last modified on	2013-03-22 15:09:37
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	10
Author	Mathprof (13753)
Entry type	Proof
Classification	msc 26D15