The smallest quadrangle inscribed in a rectangle I'm supposed to find a quadrangle of the smallest perimeter possible inscribed in a rectangle.
The inscribed quadrangle has each of its four vertices on another side of the rectangle. 
Let's call the rectangle ABCD and let the shorter side be $a$ and the larger $b$. Let's call the quadrangle $KLMN$. 
So for example $K$ lies on $AB$, $L$ on $BC$, $M$ on $CD$, $N$ on $AD$.
I think that the sides of $KLMN$ would be the shortest of its diagonals intersected at the right angle, because then by the law of cosines, we have $KL^2 + LM^2= KM^2 + 2 KL \cdot LM \cdot \cos  \angle KLM$, and $\cos  \angle KLM \le 0$ if $\angle KLM \ge 90 ^{\circ}$ and $KM$ is the shortest if it is parallel = equal to the proper side of the rectangle.
Could you tell me if I'm right or correct me if I'm wrong?
Thank you.
 A: Unflod your rectangle to make a grid and place K, L, M and N on it. You goal is to shorten the distance $KL+LM+MN+NK'$. 
You need to align $KLMNK'$ to reach the minimal distance for $KK'$. That's $2.AC$
Here is a picture of a non-optimal situation. If it were optimal, then $KLM'N'K'$ would be a straight line.

Here is a picure of an optimal situation:

A: I will prove the following: Let $ABCD$ a rectangle having $|AB|=a$ and $|BC|=c$ and let $K$, $L$, $M$, $N$ be points on $AB$, $BC$, $CD$ and $DA$, respectively. Then the perimeter of $KLMN$ is at least $2 \sqrt{a^2+b^2}$. This is achievable by letting $K$, $L$, $M$ and $N$ the midpoints of the sides of $ABCD$.
Write $k = |KB|$, $l = |LC|$, $m = |MD|$, $n = |NA|$. Then the perimeter of $KLMN$ equals $\sqrt{k^2+(b-l)^2} + \sqrt{l^2 + (a-m)^2} + \sqrt{m^2+(b-n)^2} + \sqrt{n^2+(a-k)^2}$.
We use Minkowski's inequality
$$\left({\sum_{k \mathop = 1}^n \left({a_k + b_k + c_k + d_k}\right)^p}\right)^{1/p} \le \left({\sum_{k \mathop = 1}^n a_k^p}\right)^{1/p} + \left({\sum_{k \mathop = 1}^n b_k^p}\right)^{1/p} + \left({\sum_{k \mathop = 1}^n c_k^p}\right)^{1/p} + \left({\sum_{k \mathop = 1}^n d_k^p}\right)^{1/p}$$ for $p=n=2$ and sequences $(k,b-l)$, $(a-m,l)$, $(m,b-n)$ and $(a-k,n)$.
We then find 
$$\sqrt{k^2+(b-l)^2} + \sqrt{l^2 + (a-m)^2} + \sqrt{m^2+(b-n)^2} + \sqrt{n^2+(a-k)^2} \geq \sqrt{(k+(a-m)+m+(a-k))^2+((b-l)+l+(b-n)+n)^2} = 2 \sqrt{a^2+b^2}.$$ The proof is complete. 
