Question #3ccba

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Question #3ccba

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Answer 1 · 2016-12-18T08:05:51

In order to prove this i am assuming that s is the semiperimeter of the triangle #

$s = \frac{a + b + c}{2}$ $s= (a+b+c)/2$

ok so first let

$s - a = x$ $s-a = x$
$s - b = y$ $s-b =y$
$s - c = z$ $s-c= z$

on solving for $a, b, c$ $a,b,c$ we get

$a = y + z$ $a=y+z$
$b = x + z$ $b=x+z$
$c = x + y$ $c=x+y$

Now , $\frac{a b c}{8} = \frac{(x + y) (y + z) (z + x)}{8}$ $(abc)/8 = ((x+y)(y+z)(z+x))/8$

and we know that
$\frac{x + y}{2} \geq \sqrt{x y}$ $(x+y)/2 >= sqrt(xy)$
As Arithmetic mean of $x, y$ $x,y$ is greater than their geometric mean

so ,
$\frac{(y + z) (x + y) (z + x)}{8} \geq \frac{(2 \sqrt{x y}) (2 \sqrt{y z}) (2 \sqrt{z x})}{8} = x y z$ $((y+z)(x+y)(z+x))/8 >=( (2 sqrt(x y) ) (2 sqrt(yz) )(2 sqrt(zx)))/8 = xyz$
$and x y z = (s - a) (s - b) (s - c)$ $and xyz=(s-a)(s-b)(s-c)$

Hence

$\frac{a b c}{8} \geq (s - a) (s - b) (s - c)$ $(abc)/8 >= (s-a)(s-b)(s-c)$

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Question #3ccba

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