$2\sqrt3 h\geq$ perimeter Problem:
Let $ABC$ be an acute triangle and $h$ be the longest height. Show that $2\sqrt3h\geq$ the perimeter of $ABC$.

I have a solution using geometric arguments, where I reduce the problem into isoceles triangle and bash out the length.

Here is my problem: When I tried to use algebraic approaches, I stuck at the following:
$$3(b+c-a)(c+a-b)(a+b-c)\geq c^2(a+b+c)$$
where $a,b,c$ is the side length with $c$ the minimum.
I wonder if this is doable, and if so, how? Thanks a lot.

Not entirely complete but even I've tried and I am also stuck at an inequality, I took $a$ as the smallest side
$h = \frac{2\triangle}{a} \ \  \  and \ \ 
4R\triangle = abc\\
h = \frac{bc}{2R}\\
h = \frac{bc\ sina}{a}\\
$
According to the question
$$\ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ 2\sqrt3h > 2s \\\\
 \ \ \ \ \ \frac{\sqrt3bcsina}{a} > s \\ 
 \ \ \ \ \ \ \ \ \ \  \ \ \ \ \  \ \ \ \ \  \ sin(a) > \frac{as}{\sqrt3 bc}\\
 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ sin^2(a) > \frac{a^2s^2}{3b^2c^2}\\
 \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ 1 - cos^2(a) > \frac{a^2s^2}{3b^2c^2}\\
1 - \frac{(b^2 + c^2 - a^2)^2}{4b^2c^2} > \frac{a^2s^2}{3b^2c^2}\\ 
12b^2c^2 - 3(b^2 + c^2 - a^2)^2 > 4a^2s^2 \\
 \ \ \ \ \ \ \ \ \ \ 12b^2c^2 - 3(b^2 + c^2 - a^2)^2 > a^2(a+b+c)^2
$$
Simplyfing it is
$6\displaystyle \sum_{cyc}a^2 b^2 > 3(\displaystyle \sum_{cyc}a^4) + a^2(\displaystyle \sum_{cyc}a^2) + 2a^2\displaystyle \sum_{cyc}(ab)$
Objective is to prove $a<b$ and $a<c$ I suppose but I've had no luck so far, I'll comment if I've managed to find something.
