# Where am I going wrong in this AM-GM ineguality problem?

Q: When is the perimeter of a triangle minimum for a given area?
This is how I went about it:

Let the sides of the triangle be $a$, $b$ & $c$.

$s = \dfrac{a+b+c}{2}$
By AM-GM inequality,
$\dfrac{s+(s-a)+(s-b)+(s-c)}{4}\ge \sqrt[4]{s(s-a)(s-b)(s-c)}$ ---$({1})$
Since $\sqrt[4]{s(s-a)(s-b)(s-c)}$ is constant(say $k$),
Therefore, min. value of $\dfrac{s+(s-a)+(s-b)+(s-c)}{4}$ is $k$.
i.e. min. value of $a+b+c$ is $4k$
But for this equality must hold in $(1)$, which is only possible when
$s = (s-a) = (s-b) = (s-c)$
But this implies that $a=b=c=0$ which is surely not true.

So, can you point out where precisely the error is? Thanks in advance.

• $\dfrac{s+(s-a)+(s-b)+(s-c)}{4}\ge \sqrt[4]{s(s-a)(s-b)(s-c)}$, not $\le$. – user236182 Oct 1 '15 at 13:39
• There is no error. The value $4k$ is unachievable, therefore it's not the minimum. You've successfully found that $a+b+c>4k$. – user236182 Oct 2 '15 at 13:18
• s depends on a, b, and c. (s+(s-a)+...)/4 = (4s-2s)/4 = s/2. – marty cohen Oct 2 '15 at 17:49
• As far as I know, the perimeter of a triangle is minimum for a given area when the triangle in equilateral.!? – Rishabh_Ranjan Oct 3 '15 at 10:02
• @user236182 the value of k is not unreachable. k is easily calculated since the area is given as a parameter and therefore $\sqrt{s(s-a)(s-b)(s-c)}$ is known. – Rishabh_Ranjan Oct 3 '15 at 10:05

Someone's been pulling your leg - this is a trick question. The perimeter of a triangle of a given area $S$ is unbounded! Indeed, for any $P>0$, we can construct a triangle whose area is $S$ and whose perimeter is greater than $P$.
Consider a triangle whose base is of length $P$ and whose height is $\frac{2 S}{P}$. Its area is $$\frac{height \cdot base}{2} = \frac{\frac{2 S}{P} \cdot P}{2} = S.$$ But since it has an edge of length $P$, its perimeter is surely greater than $P$.
Consider a right triangle with sides $n, 1/n, \sqrt{n^2+1/n^2}$. This has area $\frac12$ and perimeter greater than $2n$.