# 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 ProofOfArithmeticgeometricharmonicMeansInequality 2013-03-22 15:09:37 2013-03-22 15:09:37 Mathprof (13753) Mathprof (13753) 10 Mathprof (13753) Proof msc 26D15