Question #3ccba

1 Answer
Dec 18, 2016

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

s= (a+b+c)/2

ok so first let

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

on solving for a,b,c we get

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

Now , (abc)/8 = ((x+y)(y+z)(z+x))/8

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

so ,
((y+z)(x+y)(z+x))/8 >=( (2 sqrt(x y) ) (2 sqrt(yz) )(2 sqrt(zx)))/8 = xyz
and xyz=(s-a)(s-b)(s-c)

Hence

(abc)/8 >= (s-a)(s-b)(s-c)