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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.

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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.

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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 $\frac{1}{3\sqrt{3}}$.

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