# Weizenbock’s inequality

In a triangle^{} with sides $a$, $b$, $c$, and with area $A$, the following inequality holds:

$${a}^{2}+{b}^{2}+{c}^{2}\ge 4A\sqrt{3}$$ |

The proof goes like this: if $s=\frac{a+b+c}{2}$ is the semiperimeter of the
triangle, then from Heron’s formula^{} we have:

$$A=\sqrt{s(s-a)(s-b)(s-c)}$$ |

But by squaring the latter and expanding the parentheses we obtain:

$$16{A}^{2}=2({a}^{2}{b}^{2}+{a}^{2}{c}^{2}+{b}^{2}{c}^{2})-({a}^{4}+{b}^{4}+{c}^{4})$$ |

Thus, we only have to prove that:

$${({a}^{2}+{b}^{2}+{c}^{2})}^{2}\ge 3[2({a}^{2}{b}^{2}+{a}^{2}{c}^{2}+{b}^{2}{c}^{2})-({a}^{4}+{b}^{4}+{c}^{4})]$$ |

or equivalently:

$$4({a}^{4}+{b}^{4}+{c}^{4})\ge 4({a}^{2}{b}^{2}+{a}^{2}{c}^{2}+{b}^{2}{c}^{2})$$ |

which is trivially equivalent^{} to:

$${({a}^{2}-{b}^{2})}^{2}+{({a}^{2}-{c}^{2})}^{2}+{({b}^{2}-{c}^{2})}^{2}\ge 0$$ |

Equality is achieved if and only if $a=b=c$ (i.e. when the triangle is equilateral) .

See also the Hadwiger-Finsler inequality, from which this result follows as a corollary.

Title | Weizenbock’s inequality |
---|---|

Canonical name | WeizenbocksInequality |

Date of creation | 2013-03-22 13:19:33 |

Last modified on | 2013-03-22 13:19:33 |

Owner | mathcam (2727) |

Last modified by | mathcam (2727) |

Numerical id | 9 |

Author | mathcam (2727) |

Entry type | Theorem |

Classification | msc 51F99 |

Related topic | HadwigerFinslerInequality |