frusciante said:
Currently revising for my A-Level maths (UK), there is unfortunately no key in the book;
Given the triangle with sides a,b,c respectively and the area S, show that ab+bc+ca => 4*sqrt(3)*S
I have tried using the Ravi transformation without luck, any takers?
I see that if we take the area of the triangle using the Heron's formula, which is $$S=\frac{\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}}{4}$$, and then let $$x=a+b-c,\,y=b+c-a,\,z=a+c-b$$, the problem can be turned into a rather straightforward inequality problem...
We're asked to prove
$$\begin{align*}ab+bc+ca &\ge 4\sqrt{3}\frac{\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}}{4}\\&=\sqrt{3(a+b+c)(a+b-c)(b+c-a)(c+a-b)}\end{align*}$$
After the substitution, expansion and simplification, we see that we need to prove:
$$(x+y)(x+z)+(x+y)(x+z)+(x+y)(x+z)\ge 4\sqrt{3xyz(x+y+z)}$$
$$x^2+y^2+z^2+3xy+3yz+3zx\ge 4\sqrt{3xyz(x+y+z)}$$
But since $$x^2+y^2+z^2\ge xy+yz+zx$$ (from the Rearrangement or AM-GM inequality), we have:
$$x^2+y^2+z^2+3xy+3yz+3zx\ge xy+yz+zx+3xy+3yz+3zx=4xy+4yz+4zx$$
I will leave it to you to prove why $$4xy+4yz+4zx \ge 4\sqrt{3xyz(x+y+z)}$$ must hold, and please take note that my proposed solution isn't the only way to tackle the problem.(Smile)
