Question

Question #3ccba

Answer 1

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) #