Salix alba: Category:Contact geometry

2011-11-15T10:43:38Z

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{{distinguish|Weitzenböck identity}}

[[Image:LabeledTriangle.svg|thumb|right|220px|According to Weitzenböck's inequality, the [[area]] of this [[triangle]] is at most {{math|(''a''2 + ''b''2 + ''c''2) ⁄ 4√{{overline|3}}.}}]]
In [[mathematics]], '''Weitzenböck's inequality''', named after [[Roland Weitzenböck]], states that for a triangle of side lengths <math>a</math>, <math>b</math>, <math>c</math>, and area <math>\Delta</math>, the following inequality holds:

: <math>a^2 + b^2 + c^2 \geq 4\sqrt{3}\, \Delta. </math>

Equality occurs if and only if the triangle is equilateral. [[Pedoe's inequality]] is a generalization of Weitzenböck's inequality.

== Proofs ==
The proof of this inequality was set as a question in the [[International Mathematical Olympiad]] of 1961. Even so, the result is not too difficult to derive using [[Heron's formula]] for the area of a triangle:

:<math>
\begin{align}
\Delta & {} = \frac{\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}}{4} \\
& {} = \frac{1}{4} \sqrt{2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4)}.
\end{align}
</math>

=== First method ===
This method assumes no knowledge of inequalities except that all squares are nonnegative.

: <math>
\begin{align}
{} & (a^2 - b^2)^2 + (b^2 - c^2)^2 + (c^2 - a^2)^2 \geq 0 \\
{} \iff & 2(a^4+b^4+c^4) - 2(a^2 b^2+a^2c^2+b^2c^2) \geq 0 \\
{} \iff & \frac{4(a^4+b^4+c^4)}{3} \geq \frac{4(a^2 b^2+a^2c^2+b^2c^2)}{3} \\
{} \iff & \frac{(a^4+b^4+c^4) + 2(a^2 b^2+a^2c^2+b^2c^2)}{3} \geq 2(a^2 b^2+a^2c^2+b^2c^2)-(a^4+b^4+c^4) \\
{} \iff & \frac{(a^2 + b^2 + c^2)^2}{3} \geq (4\Delta)^2,
\end{align}
</math>

and the result follows immediately by taking the positive square root of both sides. From the first inequality we can also see that equality occurs only when <math>a = b = c</math> and the triangle is equilateral.

=== Second method ===
This proof assumes knowledge of the [[rearrangement inequality]] and the [[arithmetic-geometric mean inequality]].

: <math>
\begin{align}
& & a^2 + b^2 + c^2 & \geq & & ab+bc+ca \\
\iff & & 3(a^2 + b^2 + c^2) & \geq & & (a + b + c)^2 \\
\iff & & a^2 + b^2 + c^2 & \geq & & \sqrt{3 (a+b+c)\left(\frac{a+b+c}{3}\right)^3} \\
\iff & & a^2 + b^2 + c^2 & \geq & & \sqrt{3 (a+b+c)(-a+b+c)(a-b+c)(a+b-c)} \\
\iff & & a^2 + b^2 + c^2 & \geq & & 4 \sqrt3 \Delta.
\end{align}
</math>

As we have used the rearrangement inequality and the arithmetic-geometric mean inequality, equality only occurs when <math>a = b = c</math> and the triangle is equilateral.

=== Third method ===

It can be shown that the area of the inner [[Napoleon's triangle]] is:

: <math>\frac{1}{6}(a^2 + b^2 + c^2 - 4\sqrt{3}\, \Delta) </math>

and therefore greater than or equal to 0.

== External links ==
*{{MathWorld | urlname=WeitzenboecksInequality | title=Weitzenböck's Inequality}}
*"[http://demonstrations.wolfram.com/WeitzenboecksInequality/ Weitzenböck's Inequality]," an interactive demonstration by Jay Warendorff - [[Wolfram Demonstrations Project]].

{{DEFAULTSORT:Weitzenbock's inequality}}
[[Category:Elementary geometry]]
[[Category:Geometric inequalities]]
[[Category:Triangle geometry]]
[[Category:Articles containing proofs]]

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Salix alba: Category:Contact geometry