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

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.

• Math rewards you when you respect the symmetries you're given. The left hand side treats each side of the triangle equally (there is, for instance, no distinguished "base" side), so it would be nice if you could express the area of the triangle the same way. Have you heard about Heron's formula? May 3, 2017 at 7:48
• forumgeom.fau.edu/FG2005volume5/FG200519.pdf May 3, 2017 at 7:56
• This question is duplicate. There is already an answer. math.stackexchange.com/questions/2262001/… May 3, 2017 at 10:29
• Sep 24, 2018 at 10:46

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

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!

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.