Question

ABCD is quadrilateral. Is AB + BC + CD + DA >AC + BD?

Answer 1

In a triangle, the sum of the lengths of either two sides is always greater than the third side.
Considering $ΔABC$,
$AB + BC > CA$ (i)
In $ΔBCD$,
$BC + CD > DB$ (ii)
In $ΔCDA$,
$CD + DA > AC$ (iii)
In $ΔDAB$,
$DA + AB > DB$ (iv)
Adding equations (i), (ii), (iii), and (iv), we obtain
$AB + BC + BC + CD + CD + DA + DA + AB > AC + BD + AC + BD$
$\Rightarrow 2AB + 2BC + 2CD +2DA > 2AC + 2BD$
$\Rightarrow2(AB + BC + CD + DA) > 2(AC + BD)$
$\Rightarrow(AB + BC + CD + DA) > (AC + BD)$ 

Yes, the given expression is true.