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 first way does not give a solution because the equality
$$s=s-a=s-b=s-c$$ is impossible.
By the way, by AM-GM
$$\frac{A}{s^2}=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s^2}\leq\frac{\sqrt{s\left(\frac{s-a+s-b+s-c}{3}\right)^3}}{s^2}=\frac{1}{3\sqrt3}.$$
The equality occurs for $$s-a=s-b=s-c$$ or $$a=b=c,$$ which says that we got a maximal value.
