Among rectangles of equal area, which one has the smallest perimeter?
We denote the sides of the rectangle by $X$ and $Y$. Then,

$$
T = x y, \quad K = 2(x + y)
$$

We know that if $x \neq y$, then

$$
\frac{x + y}{2} > \sqrt{x y}
$$

which means

$$
\frac{K}{4} > \sqrt{T}, \quad K > 4 \sqrt{T}
$$

The smallest possible value of $K$ is therefore $4 \sqrt{T}$, which is only achieved when $x = y = \sqrt{T}$, i.e., when the rectangle is a square.

Solved by; Reichlin V., Tar D., Zatykó L.

Using differential calculus: Villányi O.