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If Delta is the area of a triangle with side lengths a,b,c then show that Delta<=1/4sqrt((a+b+c)*abc). Also show that equality occurs if and only if a=b=c.

1 Answer
Jun 6, 2018

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