Here, ABCD is a quadrilateral and AC and BD are its diagonals.

Now, As we that, sum of two sides of a triangle is greater than the third side.

∴ In Δ ACB,

AB + BC > AC (i)

In Δ BDC,

CD + BC > BD (ii)

In Δ BAD,

AB + AD>BD (iii)

In Δ ACD,

AD + DC > AC (iv)

Now, adding (i), (ii), (iii) and (iv):

AB + BC + CD + BC + AB + AD + AD + DC> AC + BD + BD + AC

2AB + 2BC + 2CD + 2AD > 2AC + 2BD

Thus, AB + BC + CD + AD > AC + BD

Hence, proved