Problem. Let ABCD be a rhombus with < BAD = 60°. The points S and R lie inside the triangles ABD and DBC respectively such that < SBR = < RDS = 60°. Prove that $SR^2 \geq AS \cdot CR$. Solution by Myth and me. We start with a simple lemma: Lemma 1. For any angle t, we have $\sin\left( 60^{\circ}-t\right)+\sin\left( 60^{\circ}+t\right) \leq2\sin60^{\circ}$. Proof. Using the addition formulas for sin, we get: $\sin\left( 60^{\circ}-t\right) +\sin\left( 60^{\circ}+t\right)$ $=\left( \sin60^{\circ}\cos t-\cos60^{\circ}\sin t\right) +\left(\sin60^{\circ}\cos t+\cos60^{\circ}\sin t\right)$ $=2\sin60^{\circ}\cos t\leq2\sin60^{\circ}$ (since $\sin 60^{\circ}=\frac{\sqrt3}{2}\geq 0$ and $\cos t\leq 1$). Now let's solve the problem: Since the quadrilateral ABCD is a rhombus, we have DA = AB. Since we also have < BAD = 60°, the triangle BAD is equilateral; since a rhombus is symmetric with respect to each of its diagonals, the triangles BAD and BCD are congruent, so that the triangle BCD is also equilateral. WLOG assume that $DS\leq DR$. Let X be a point on the ray DS such that DX = DR, and let Y be a point on the ray DR such that DY = DS. Then, since DX = DR and < RDX = 60° (this follows from < RDS = 60°), the triangle XDR is equilateral. Similarly, the triangle YDS is equilateral. Since the triangle XDR is equilateral, we have < DXR = 60°. In other words, < SXR = 60°. The equilateral triangle YDS yields < DYS = 60°, so that < RYS = 180° - < DYS = 180° - 60° = 180° - < SXR. This implies that the quadrilateral RYSX is cyclic. In other words, the points R, Y, S and X lie on one circle. The point B also lies on this circle, since < SBR = 60° and < SXR = 60° imply < SBR = < SXR. Hence, all five points R, Y, S, X and B lie on one circle. Let r be the radius of this circle. Then, by the fact that the length of a chord in a circle equals the double circumradius times the sine of the chordal angle of this chord, we have $SR=2r\sin\measuredangle SBR$, $BX=2r\sin\measuredangle BSX$ and $BY=2r\sin\measuredangle BSY$. Since the triangle YDS is equilateral, we have < DSY = 60°, and thus < BSX + < BSY = 180° - < DSY = 180° - 60° = 120°. Hence, there exists an angle t such that < BSX = 60° - t and < BSY = 60° + t (this angle t can be positive or negative or 0°). By Lemma 1, we thus have $\sin\measuredangle BSX+\sin\measuredangle BSY=\sin\left( 60^{\circ}-t\right)+\sin\left( 60^{\circ}+t\right) \leq2\sin60^{\circ}=2\sin\measuredangle SBR$ (since < SBR = 60°). Hence, $2r\sin\measuredangle BSX+2r\sin\measuredangle BSY\leq2\cdot2r\sin \measuredangle SBR$, or, in other words, $BX+BY\leq2\cdot SR$. Since the triangle BAD is equilateral, we have < ADB = 60°, what, combined with < RDS = 60°, yields < RDS = < ADB, so that < RDB = < RDS - < SDB = < ADB - < SDB = < SDA. In other words, < YDB = < SDA. Since we also have BD = AD (this is again because the triangle BAD is equilateral) and DY = DS, it follows that the triangles YDB and SDA are congruent, and thus BY = AS. Similarly, BX = CR. Thus, the inequality $BX+BY\leq2\cdot SR$ rewrites as $CR+AS\leq2\cdot SR$. In other words, we have $2\cdot SR\geq AS+CR$. Thus, $SR\geq\frac{AS+CR}{2}$. But by the AM-GM inequality, $\frac{AS+CR}{2}\geq\sqrt{AS\cdot CR}$. Hence, $SR\geq\sqrt{AS\cdot CR}$, and $SR^2\geq AS\cdot CR$. And we are done. Equality holds if and only if < BSD = 120°. The proof is left to the reader (hint: < BSD = 120° is equivalent to t = 0° and to AS = CR = SR).