What's the maximum of the sum of the lengths of the diagonals in a convex quadrilateral whose perimeter is 4?

2. This is trial and error -

1. A square of side 1 has sum of diagonals 2 srqt(2) = 2.8+.

2. A flattened square (think parallelogram of side 1 with 2 parallel sides both nearly on the x-axis) has 1 diagonal of length 2 and the other of length 0, so sum is 2.

3. A rectangle of sides (2-x,x,2-x,x), with x just a little bit larger than 0 has diagonals of about 2 and 2, so the sum is 4.

So I'd guess the maximum length of 2 diagonals of a quadrilateral of perimeter 4 is 4.

3. Originally Posted by qmech
This is trial and error -

1. A square of side 1 has sum of diagonals 2 srqt(2) = 2.8+.

2. A flattened square (think parallelogram of side 1 with 2 parallel sides both nearly on the x-axis) has 1 diagonal of length 2 and the other of length 0, so sum is 2.

3. A rectangle of sides (2-x,x,2-x,x), with x just a little bit larger than 0 has diagonals of about 2 and 2, so the sum is 4.

So I'd guess the maximum length of 2 diagonals of a quadrilateral of perimeter 4 is 4.
OK but it's easy to show that it cannot be more than 4 because of the triangle inequality. Just take the two triangles with one common diagonal side and write the 2 inequalities, add them and you get that $2e. This goes for the other diagonal so $e+f<4$. But I'm asked for the maximum and where it attains. As you said we can make it sufficiently close to 4 but we can't reach it. Therefore I guess there is no maximum.
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