Show that sum of all sides of a quadrilateral is greater than the sum of the diagonals

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Solution

Consider a quadrilateral $A B C D$
Consider $Δ A C D$
By triangle inequality
$A C < A D + C D \dots (1)$
Consider $Δ A C B$
By triangle inequality
$A C < A B + C B \dots (2)$
Consider $Δ B C D$
By triangle inequality
$B D < B C + C D \dots (3)$
Consider $Δ A D B$
By triangle inequality
$B D < A D + A B \dots (4)$
Adding all the equations
$2 (A B + B C + C D + A D) > 2 (A C + B D) A B + B C + C D + D A > A C + B D$

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Show that sum of all sides of a quadrilateral is greater than the sum of the diagonals

Consider a quadrilateral ABCDConsider ΔACDBy triangle inequality AC<AD+CD⋯(1)Consider ΔACBBy triangle inequality AC<AB+CB⋯(2)Consider ΔBCDBy triangle inequality BD<BC+CD⋯(3)Consider ΔADBBy triangle inequality BD<AD+AB⋯(4)Adding all the equations2(AB+BC+CD+AD)>2(AC+BD)AB+BC+CD+DA>AC+BD