Constant Perimeters

A square with perimeter $12$ has a side length of $\dfrac{12}4 = 3$ , and hence its area is $9$ .
A triangle with perimeter $12$ has sides $a,b,c$ for which $a+b+c=12$ , and hence its area is $\sqrt{6\left(6-a\right)\left(6-b\right)\left(6-c\right)}$ . ( Heron's formula )
Also, $0 < a < b + c, 0 < b < c + a, 0 < c < a + b \implies 0<a,b,c<6$ (The first one is true for sides of any triangle.)

Now using AM-GM inequality ,

$\begin{aligned}\sqrt[3]{\left(6-a\right)\left(6-b\right)\left(6-c\right)} \leq \dfrac{18-a-b-c}3 = 2 &\implies \left(6-a\right)\left(6-b\right)\left(6-c\right) \leq 8 \\ &\implies 6\left(6-a\right)\left(6-b\right)\left(6-c\right) \leq 48 < 81 \\ &\implies \sqrt{6\left(6-a\right)\left(6-b\right)\left(6-c\right)} < 9 \end{aligned}$

Hence, $\boxed{\text{A square with perimeter 12}}$ has always larger area.