The sides $a$, $b$, $c$ of a triangle with area $D$ are such that $ab+bc+ca\ge 4\sqrt{3}D$

Question

The sides $a$, $b$, $c$ of a triangle with area $D$ are such that $ab+bc+ca\ge 4\sqrt{3}D$.

I think I need to figure out a way to apply the A.M.-G.M. inequality here, but couldn't find a proper way of doing so.

Answer 1

Hint: $\sqrt{ab\sin C}+\sqrt{bc\sin A}+\sqrt{ca\sin B}=3\sqrt{2D}$. Use Cauchy-Schwarz and then maximise $\sin A+\sin B+\sin C$.

Answer 2

We need to prove that $$ab+ac+bc\geq\sqrt{3\sum_{cyc}(2a^2b^2-a^4)}$$ or $$\sum_{cyc}(3a^4-5a^2b^2+2a^2bc)\geq0$$ or $$\sum_{cyc}(6a^4-6a^2b^2-4a^2b^2+4a^2bc)\geq0$$ or $$\sum_{cyc}(a-b)^2(3(a+b)^2-2c^2)\geq0,$$ for which it's enough to prove that $$\sum_{cyc}(a-b)^2(2(a+b)^2-2c^2)\geq0$$ or $$\sum_{cyc}(a-b)^2(a+b-c)(a+b+c)\geq0,$$ which is obviously true.

Done!

Answer 3

I realise this isn't answering your question, but it was interesting so I have decided to post it anyway. tl;dr I showed that $ab+ac+bc \geq 6D$, a weaker inequality, using AM-GM.

The area of triangle (in terms of its $3$ sides) is given by: $D=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$. This is known as Heron's formula.

This means we want to show that $ab+bc+ac \geq \sqrt{3(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$

First note that \begin{align*}(a+b+c)(-a+b+c)+(a-b+c)(a+b-c) &= (-a^2+b^2+c^2+2bc)+(a^2-b^2-c^2+2bc)\\ &= 4bc\end{align*}

Hence by AM-GM, $\frac{4bc}{2} = 2bc \geq \sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)} = 4D$

By symmetry we also have $2ab \geq 4D$, $2ac \geq 4D$.

Now by adding the three equations, we have $2(ab+ac+bc) \geq 12D$, so $ab+ac+bc \geq 6D$. Unfortunately $4\sqrt{3} \approx 6.9$, so proved a weaker result.