Triangle side terms and area inequality

[frusciante]

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?

 

[anemone]

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 \(\displaystyle S=\frac{\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}}{4}\), and then let \(\displaystyle x=a+b-c,\,y=b+c-a,\,z=a+c-b\), the problem can be turned into a rather straightforward inequality problem...

We're asked to prove

\(\displaystyle \begin{align*}ab+bc+ca &\ge 4\sqrt{3}\frac{\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}}{4}\\&=\sqrt{3(a+b+c)(a+b-c)(b+c-a)(c+a-b)}\end{align*}\)

After the substitution, expansion and simplification, we see that we need to prove:

\(\displaystyle (x+y)(x+z)+(x+y)(x+z)+(x+y)(x+z)\ge 4\sqrt{3xyz(x+y+z)}\)

\(\displaystyle x^2+y^2+z^2+3xy+3yz+3zx\ge 4\sqrt{3xyz(x+y+z)}\)

But since \(\displaystyle x^2+y^2+z^2\ge xy+yz+zx\) (from the Rearrangement or AM-GM inequality), we have:

\(\displaystyle x^2+y^2+z^2+3xy+3yz+3zx\ge xy+yz+zx+3xy+3yz+3zx=4xy+4yz+4zx\)

I will leave it to you to prove why \(\displaystyle 4xy+4yz+4zx \ge 4\sqrt{3xyz(x+y+z)}\) must hold, and please take note that my proposed solution isn't the only way to tackle the problem.(Smile)

 

[Evgeny.Makarov]

This problem is discussed (with solution) in the following book.

R.B. Manfrino, J.A.G. Ortega, R.V. Delgado. Inequalities: A Mathematical Olympiad Approach. 2009. p. 74, exercise 2.53.