Prove that area of an arbitrary triangle is less than one fourth of the square of its perimeter.

(The question is from my sequences and series text)

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- May 21st 2009, 09:53 AMfardeen_genArea of an arbitrary triangle?
Prove that area of an arbitrary triangle is less than one fourth of the square of its perimeter.

(The question is from my sequences and series text) - May 21st 2009, 10:46 AMGrandad
Hello fardeen_genDo you know the formula for the area of a triangle: $\displaystyle \Delta = \sqrt{s(s-a)(s-b)(s-c)}$, where the semi-perimeter $\displaystyle s = \tfrac12(a+b+c)$?

Well, each of $\displaystyle (s-a), (s-b), (s-c)$ is less than $\displaystyle s$. So $\displaystyle \Delta < \sqrt{s^4} = \tfrac14(a+b+c)^2$.

Grandad