In a quadrilateral ABCD,

1. AB + BC + CD + DA > AC + BD

2. AB + BC + CD + DA < AC + BD

3. AB + BC + CD + DA = AC + BD

4. AB + BC < AC

We know that,

The sum of lengths of two sides of a triangle is always greater than the third side.

In △ ABC,

⇒ AB + BC > AC ……..(1)

In △ ADC,

⇒ AD + CD > AC ……..(2)

In △ ADB,

⇒ AD + AB > BD ……..(3)

In △ DCB,

⇒ DC + CB > BD ……..(4)

Adding equations (1), (2), (3) and (4) we get,

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

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

⇒ 2(AB + BC + CD + AD) > 2(AC + BD)

⇒ AB + BC + CD + AD > AC + BD.

Hence, Option 1 is the correct option.