Editing Triangle inequality (section)

===Examples of use===
Consider a triangle whose sides are in an [[arithmetic progression]] and let the sides be {{math|''a'', ''a'' + ''d'', ''a'' + 2''d''}}. Then the triangle inequality requires that

:<math>\begin{array}{rcccl} 
  0 &<& a &<& 2a+3d, \\
  0 &<& a+d &<& 2a+2d, \\
  0 &<& a+2d &<& 2a+d. 
\end{array}</math>

To satisfy all these inequalities requires

:<math> a>0 \text{ and } -\frac{a}{3}<d<a. </math><ref>{{cite journal|title=input: ''solve 0<a<2a+3d, 0<a+d<2a+2d, 0<a+2d<2a+d,'' |last=Wolfram{{!}}Alpha|journal=Wolfram Research|url=http://www.wolframalpha.com/input/?i=solve%200%3Ca%3C2a%2B3d%2C%200%3Ca%2Bd%3C2a%2B2d%2C%200%3Ca%2B2d%3C2a%2Bd&t=ff3tb01|access-date=2010-09-07}}</ref>

When {{mvar|d}} is chosen such that {{math|''d'' {{=}} ''a''/3}}, it generates a right triangle that is always similar to the [[Pythagorean triple]] with sides {{math|3}}, {{math|4}}, {{math|5}}.

Now consider a triangle whose sides are in a [[geometric progression]] and let the sides be {{math|''a'', ''ar'', ''ar''<sup>2</sup>}}. Then the triangle inequality requires that

:<math>\begin{array}{rcccl} 
  0 &<& a &<& ar+ar^2, \\
  0 &<& ar &<& a+ar^2, \\
  0 &<& \! ar^2 &<& a+ar. 
\end{array}</math>

The first inequality requires {{math|''a'' > 0}}; consequently it can be divided through and eliminated. With {{math|''a'' > 0}}, the middle inequality only requires {{math|''r'' > 0}}. This now leaves the first and third inequalities needing to satisfy

:<math>
\begin{align}
r^2+r-1 & {} >0 \\
r^2-r-1 & {} <0.
\end{align}
</math>

The first of these quadratic inequalities requires {{mvar|r}} to range in the region beyond the value of the positive root of the quadratic equation {{math|''r''<sup>2</sup> + ''r'' − 1 {{=}} 0}}, i.e. {{math|''r'' > ''φ'' − 1}}  where {{mvar|φ}} is the [[golden ratio]]. The second quadratic inequality requires {{mvar|r}} to range between 0 and the positive root of the quadratic equation {{math|''r''<sup>2</sup> − ''r'' − 1 {{=}} 0}}, i.e. {{math|0 < ''r'' < ''φ''}}. The combined requirements result in {{mvar|r}} being confined to the range
:<math>\varphi - 1 < r <\varphi\, \text{ and } a >0.</math><ref>{{cite journal|title=input: ''solve 0<a<ar+ar<sup>2</sup>, 0<ar<a+ar<sup>2</sup>, 0<ar<sup>2</sup><a+ar'' |last=Wolfram{{!}}Alpha|journal=Wolfram Research|url=http://wolframalpha.com/input?i=solve+0%3Ca%3Car%2Bar^2%2C+0%3Car%3Ca%2Bar^2%2C+0%3Car^2%3Ca%2Bar|access-date=2010-09-07}}</ref>

When {{mvar|r}} the common ratio is chosen such that {{math|''r'' {{=}} {{sqrt|''φ''}}}} it generates a right triangle that is always similar to the [[Kepler triangle]].