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# proof of Nesbitt’s inequality

Starting from Nesbitt’s inequality

$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}$ |

we transform the left hand side:

$\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3\geq\frac{3}{2}.$ |

Now this can be transformed into:

$((a+b)+(a+c)+(b+c))\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\geq 9.$ |

Division by 3 and the right factor yields:

$\frac{(a+b)+(a+c)+(b+c)}{3}\geq\frac{3}{\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b% +c}}.$ |

Now on the left we have the arithmetic mean and on the right the harmonic mean, so this inequality is true.

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## Versions

(v6) by mathwizard 2013-03-22