Let $a,b,c$ be the sides of a triangle. The find maximum value $ \frac{A}{S^{2}}$ Let $a,b,c$ be the sides of a triangle, $A$ is the area and $S$ is the semi-perimeter $(a+b+c)/2$. 
Find the maximum value $\frac{A}{S^{2}}$. 
My Approach:
Method 1:
Applying AM-GM inequality on $S,S-a,S-b,S-c$
$$\frac{4S-2S}{4} \ge \sqrt[4]{S(S-a)(S-b)(S-c)}$$
$$\frac{1}{4}\ge \frac{A}{S^2}$$
Method 2:
Applying AM-GM inequality on $S-a,S-b,S-c$
$$\frac{3S-2S}{3} \ge \sqrt[3]{(S-a)(S-b)(S-c)}$$
$$\frac{1}{3\sqrt{3}}\ge \frac{A}{S^2}$$
For equality in method 1 $a=b=c=0$ which is not true hence we get maximum value from method 2. But not sure if $\frac{1}{3\sqrt{3}}$ is the maximum because some other method may give me some other maximum value. So is there a method which gives me the exact maximum value.

Your Method 2 is fine and it gives the exact maximum value that you are looking for.
Since $A=\sqrt{S(S-a)(S-b)(S-c)}$, the AM-GM inequality in Method 2 yields
$$\frac{(S-a)+(S-b)+(S-c)}{3} \ge \sqrt[3]{(S-a)(S-b)(S-c)}$$
that is
$$\frac{S^3}{3^3}\geq(S-a)(S-b)(S-c)=\frac{A^2}{S},$$
and therefore
$$\frac{1}{3\sqrt{3}}\ge \frac{A}{S^2}.$$
Here equality holds when $S-a=S-b=S-c$ that is when $a=b=c$ and we may conclude that the maximum value of $A/S^2$ is just $\frac{1}{3\sqrt{3}}$.
