Considering the Heron's formula of area of the triangle ABC
Delta=sqrt(s(s-a)(s-b)(s-c)), where a,bandc are three sides of the triangle and its semi perimeter is s i.e.2s=a+b+c
Now let
s-a=x,s-b=yands-c=z
Now each of x,y and z is greater than zero as sum of two sides of a triangle is greater than third side.
So
x+y=2s-a-b=a+b+c-a-b=c
Similarly
y+z=2s-b-c=a
And
z+x=2s-c-a=b
Now we know AM>GM
So (x+y)/2>=sqrt(xy)
(y+z)/2>=sqrt(yz)
(z+x)/2>=sqrt(zx)
So multiplying we get
sqrt(xy*yz*zx) <= 1/8(x+y)(y+z)(z+x)
=>sqrt(x^2y^2z^2)<=1/8(x+y)(y+z)(z+x)
=>xyz<=1/8(x+y)(y+z)(z+x)
=>(s-a)(s-b)(s-c)<=1/8cba
=>s(s-a)(s-b)(s-c)<=1/8scba
=>s(s-a)(s-b)(s-c)<=1/8*1/2(a+b+c)abc
=>sqrt(s(s-a)(s-b)(s-c))<=sqrt(1/16(a+b+c)abc)
=>Delta<=1/4sqrt((a+b+c)abc)
We know for an equilateral triangle of side a
Delta=sqrt3/4a^2. So equality of the above relation is satisfied when a=b=c