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# Cyclic Inequalities

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. $$\Box$$

- 3 years, 9 months ago

Good job!

- 3 years, 9 months ago