# 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 $$S,S-a,S-b,S-c$$

$$\frac{4S-2S}{4} \ge \sqrt{S(S-a)(S-b)(S-c)}$$ $$\frac{1}{4}\ge \frac{A}{S^2}$$

Method 2:

Applying AM-GM inequality on $$S-a,S-b,S-c$$

$$\frac{3S-2S}{3} \ge \sqrt{(S-a)(S-b)(S-c)}$$ $$\frac{1}{3\sqrt{3}}\ge \frac{A}{S^2}$$

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.

## 2 Answers

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.

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 $$\frac{(S-a)+(S-b)+(S-c)}{3} \ge \sqrt{(S-a)(S-b)(S-c)}$$ 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 $$\frac{1}{3\sqrt{3}}$$.