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I need some help proving Pedoe's Inequality for two triangles, which states that

[tex]a_{1}^2(b_{2}^2+c_{2}^2-a_{2}^2) + b_{1}^2(c_{2}^2 + a_{2}^2 - b_{2}^2) + c_{1}^2(a_{2}^2 + b_{2}^2 - c_{2}^2) \geq 16F_{1}F_{2}[/tex]

where [itex](a_{1},b_{1},c_{1})[/itex] and [itex](a_{2},b_{2},c_{2})[/itex] are the sides of triangles [itex]A_{1}B_{1}C_{1}[/itex] and [itex]A_{2}B_{2}C_{2}[/itex] respectively and [itex]F_{1}[/itex], [itex]F_{2}[/itex] are their areas.

The expressions in the brackets suggest usage of the cosine rule, which gives [itex]b_{2}^2 + c_{2}^2 - a_{2}^2 = 2b_{2}c_{2}\cos A_{2}[/itex]. Using this the left hand side transforms to three terms of the type [itex]2a_{1}^2b_{2}c_{2}\cos A_{2}[/itex] but this doesn't seem to help. The right hand side can be transformed using Hero's formula for the area of either triangle. This also gets rid of 16. But I don't know how to proceed further.

I would be grateful if someone could suggest a way out. In case there is a proof available on the internet, please let me know...I am searching for it myself on google right now....so far I have found several pages just listing the theorem's statement (mostly copied from wiki).

Thanks...

Cheers

vivek

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# Pedoe's Two Triangle Inequality

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