Question: **Q:** *When is the perimeter of a triangle minimum for a given area?*  

This is how I went about it:  

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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.
> 
> 
> 

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

Answer: 
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
$2n$.

However,
on a sphere,
the perimeter and the area of triangles
are bounded.