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#