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.