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Prove that the geometric mean of the lengths of the two diagonals of a cyclic quadrilateral is always greater than the geometric mean of the lengths of the four sides.

Note by Tristan Shin 3 years, 9 months ago

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Let \(D_1,D_2\) be the diagonals and \(a,b,c,d\) be the sides.

We know that \(D_1D_2=ac+bd\), Ptolemy's Theorem. By AM-GM on the right side, we have that \(D_1D_2 = ac+bd\ge 2\sqrt{abcd} > \sqrt{abcd}\).

Thus, \(D_1D_2 > \sqrt{abcd}\), and taking the square root on both sides gives \(\sqrt{D_1D_2} > \sqrt[4]{abcd}\), as desired. \(\Box\)

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TopNewestLet \(D_1,D_2\) be the diagonals and \(a,b,c,d\) be the sides.

We know that \(D_1D_2=ac+bd\), Ptolemy's Theorem. By AM-GM on the right side, we have that \(D_1D_2 = ac+bd\ge 2\sqrt{abcd} > \sqrt{abcd}\).

Thus, \(D_1D_2 > \sqrt{abcd}\), and taking the square root on both sides gives \(\sqrt{D_1D_2} > \sqrt[4]{abcd}\), as desired. \(\Box\)

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Good job!

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