# An Inequality Not in Triangle

### Solution

With the reference to the above diagram, $OA=a,\,$ $OB=b,\,$ $OC=c,\,$ $OD=d.\,$ By the Law of Cosines,

$AB^2=a^2b^2-2ab\cos 45^{\circ}=a^2+b^2-ab\sqrt{2},\\ BC^2=b^2+c^2-2bc\cos 30^{\circ}=b^2+c^2-bc\sqrt{3},\\ CD^2=c^2+d^2-2cd\cos 15^{\circ}=c^2+d^2-\displaystyle\frac{cd(\sqrt{6}+\sqrt{2})}{2},\\ AD^2-a^2+d^2,\\ AB+BC+CD\ge AD$

and the required inequality follows.

### Acknowledgment

This is a problem and solution from the Romanian Mathematical Magazine kindly communicated to me by Dan Sitaru.