Triangular Inequality: Want resolution of the following discrepancy? There is this following Multi-correct Answer type question in my book. It asks:

If $A$ is the area and $2s$ the sum of the sides of a triangle then:
  
  
*
  
*$A \leq \frac {s^{2}}{4}$
  
*$A \leq \frac  {s^{2}}{3\sqrt 3}$
  
*$A \lt \frac {s^{2}}{\sqrt 3}$
  
*None of these

All over the internet, atleast in India, they are saying that option (1) and (2) are correct. But as far as I think the option (2) and (3) are correct whereas (1) is not. Why? Because I found that in (2), via application  of $\text{AM} \geq \text{GM}$, both equality and inequality parts are correct and hence the strict inequality of (3) is also correct. But equality of the (1) is not true as then the quantities on which the $\text{AM} \geq \text{GM}$ is applied are themselves zero ( that is to say that for the equality to be true the triangle must have zero area) whereas I have learnt that this method is only true for positive numbers. 
Now the above-mentioned blah-blah is quite contrary to what is being told on the internet. Hence I would like yourguidance/opinion  on how to proceed further to determine the answer. Also please mention any conceptual mistakes make by me.

Edit
After reading the answers here I reach the conclusion that options (1),(2) and (3) are correct.
 A: It is a rather easy matter to prove that for a triangle of given base and height, the configuration that gives the shortest perimeter is isosceles. Hence for a given area, the shortest perimeter is that of the equilateral triangle.
$$A=\frac{\sqrt 3}4c^2,\\2s=3c$$ and
$$\frac A{s^2}=\frac 1{3\sqrt3}$$ is the largest possible ratio.
A: For any given triangle, all three inequalities hold. The reason is that $(2)$ holds, and therefore $(1)$ and $(3)$ hold as well because $$\frac{1}{3\sqrt{3}}<\frac{1}{4}<\frac{1}{\sqrt{3}}.$$ Your mistake is in thinking that the assertion $p\le q$ includes the assertion that there exist a case in which equality holds, which is not true. The statement $p\le q$ just means that $p$ is always less than or equal to $q,$ but that does not means that it is necessarily possible that $p=q.$
For completeness, I will prove that $(2)$ holds. One side of Gerretsen's inequalities states that $$16Rr-5r^2 \le s^2$$ where $R$ is the circumradius, $r$ is the inradius and $s$ is the semiperimeter. Euler's inequality states that $2r\le R.$ Equality in holds in each inequality individually if and only if the triangle is equilateral.
Let $c=3\sqrt{3}.$ Then Euler is equivalent to $$\left(\frac{c^2+5}{16}\right)r\le R,$$ which in turn is equivalent to $$c^2 r^2 \le 16Rr - 5r^2.$$ By Gerretsen, $$c^2 r^2 \le s^2,$$ and so $cr\le s.$ Using the fact that $rs=A,$ we get $$A\le \frac{s^2}{c},$$ with equality holding if and only if the triangle is equilateral.
