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Homeproof of arithmetic-geometric-harmonic means inequality

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# 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 $x_{n}\geq 1\geq x_{1}$.

## Mathematics Subject Classification

26D15*no label found*

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