$A B C D$ is a quadrilateral.Prove $A B + B C + C D + D A > A C + B D$ ?

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Solution

$A B C D$ is a quadrilateral and $A C$ , and $B D$ are the diagonals.
Sum of the two sides of a triangle is greater than the third side.
So, considering the triangle $A B C, B C D, C A D$ and $B A D$ , we get
In $△ A B C, A B + B C > A C$ ......(1)
In $△ A C D, C D + A D > A C$ ......(2)
In $△ A B D, A B + A D > B D$ ......(3)
In $△ B C D, B C + C D > B D$ ......(4)
Adding equation (1), (2), (3) and ($), we get
$2 (A B + B C + C A + A D) > 2 (A C + B D)$
$2 (A B + B C + C A + A D) > 2 (A C + B D)$
$(A B + B C + C A + A D) > (A C + B D)$
Hence, proved.

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