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Posted: Fri Sep 21, 2012 9:38 am
On every side of a square with sides measuring one,choose one point. The four points will form a quadrilateral of perimeter $h$. Prove that $2\sqrt{2}\leq h$.

Posted: Sat Sep 22, 2012 1:24 pm
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Posted: Sun Sep 23, 2012 10:30 am
suppose, we want to get the quad having the shortest perimeter.
what is we can make two of the sides $0$ that will be the shortest. because, straight line is the shortest distance between two points.
so, let the $4$ vertices of the quad be on the $2$ points diagonally opposite vertices of the square.
$h=2 \sqrt2$
so, $2\sqrt2 \ngtr h$

Posted: Fri Sep 28, 2012 12:27 pm
We will use a simple lemma to prove it.
Lemma: In any right angled triangle with the sides $a,b,c$ where $c$ is a hypotenuse.Then $\sqrt{2}c \geq a+b$.(Prove the lemma yourself by $A.M-G.M$.)
Now applying it in $\Delta EBF$, we get, $\sqrt{2}EF \geq EB+BF$.
Likewise,$\sqrt{2}FG \geq FC+CG$,$\sqrt{2}HG \geq HD+GD$,$\sqrt{2}EH \geq AE+AH$.(look the figure above).Summing these inequalities we get what we wanted.

Lemma: In any right angled triangle with the sides $a,b,c$ where $c$ is a hypotenuse.Then $\sqrt{2}c \geq a+b$.(Prove the lemma yourself by $A.M-G.M$.)
Now applying it in $\Delta EBF$, we get, $\sqrt{2}EF \geq EB+BF$.
Likewise,$\sqrt{2}FG \geq FC+CG$,$\sqrt{2}HG \geq HD+GD$,$\sqrt{2}EH \geq AE+AH$.(look the figure above).Summing these inequalities we get what we wanted.