In a quadrilateral $A B C D$ ,prove that $A B + B C + C D + D A > A C + B D$ ?

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Solution

Ref.Image.
By In equality theorem, sum of any two
sides of $Δ$ is grater than third one
In $Δ$ ABC
AC $∠$ AB+BC ... (1)
In $Δ$ BCD
DB $∠$ BC +CD ... (2)
In $Δ$ ACD
AC $∠$ DA +DC ... (3)
In $Δ$ ADB
DB $∠$ AD + AB ... (4)
adding (1),(2),(3),(4)
= 2AC + 2DB $∠$ 2AB + 2BC + 2CD + 2DA
= AC+DB $∠$ AB + BC + CD +DA
Hence proved.

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