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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= (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) #