Question

ABCD is quadrilateral. Is AB + BC + CD + DA < 2 (AC + BD)?

Answer 1

In a triangle, the sum of the lengths of either two sides is always greater than the third side.
Considering $ΔOAB$,
$OA + OB > AB$ (i)
In $ΔOBC$,
$OB + OC > BC$ (ii)
In $ΔOCD$,
$OC + OD > CD$ (iii)
In $ΔODA$,
$OD + OA > DA$ (iv)

Adding equations (i), (ii), (iii), and (iv), we obtain

$OA + OB + OB + OC + OC + OD + OD + OA > AB + BC + CD + DA$
$\Rightarrow$ $2\;OA + 2\;OB + 2\;OC + 2\;OD > AB + BC + CD + DA$
$\Rightarrow$ $2\;OA + 2\;OC + 2\;OB + 2\;OD > AB + BC + CD + DA$
$\Rightarrow$ $2(OA + OC) + 2(OB + OD) > AB + BC + CD + DA$
$\Rightarrow$ $2(AC) + 2(BD) > AB + BC + CD + DA$
$\Rightarrow$ $2(AC + BD) > AB + BC + CD + DA$

Yes, the given expression is true.