Given volume of cone equal to **V**, i need to minimize side surface area of given cone **without using derivatives**. Exactly i'm trying to use **AM-GM inequality.**

First of all i tried to do it like it was done here: [Maximise right circular cone volume with fixed surface area using inequalites](https://math.stackexchange.com/questions/3133321), but got stucked - case in link is inverse:given *surface area* and we need to minimize *volume*. In addition in url OP works with **whole** surface of a cone(i need only side surface).
Here are my calculations:

Given $V; V\_{cone} = \frac{1}{3}\pi R^2 h; A = \pi R l, A\rightarrow min. $ Since $l = \sqrt{h^2 + R^2} \;and\; h = \frac{3V}{\pi R^2}:$
$$l = \sqrt{ \frac{3V}{\pi R^2} + R^2} \Rightarrow $$
$$\Rightarrow A = \pi R \sqrt{ \frac{3V}{\pi R^2} +R^2} = \pi \sqrt{ \frac{3V}{\pi} +R^4}$$
And then i got stucked. Any ideas how to solve it?

**UPD:**
I've also expressed $R$ on $h$ value. Then:
$R = \sqrt{\frac{3V}{\pi h} }, \; l = \sqrt{\frac{3V}{\pi h} + h^2 } \Rightarrow $
$$\Rightarrow A = \pi \sqrt{\frac{3V}{\pi h} } \sqrt{\frac{3V}{\pi h} + h^2 }.$$
So, 
$$A = \pi \sqrt{\frac{9V^2}{\pi^2 h^2} + \frac{3Vh}{\pi} } = \pi \sqrt{\frac{3V}{\pi}\Big(\frac{3V}{\pi h^2} +h \Big) }\rightarrow \min. $$
Hence $ \min is \; achieving \; \iff $ $h = -\frac{3V}{\pi h^2} \Rightarrow h^3 = -\frac{3V}{\pi}.$ 

**BUT** A is equal to zero in that case. Will it be the right solution?