<s>[INST] 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.
 [/INST] 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.
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