Maximum value

Let $ABC$ be a triangle. Find the maximum value of:

$$R=\sqrt{2}\sin A + \sqrt{5} \sin B +\sqrt{10}\sin C$$

Solution



$$\begin{aligned}
R &=\sqrt{2}\sin A +\sqrt{5}\sin B +\sqrt{10}\sin C \\
 &= \sqrt{2} \sin A + 1\cdot \sqrt{5}\sin B +\sin A \cdot \sqrt{10}\cos B + \sqrt{2}\cos A \cdot \sqrt{5}\sin B\\
 &\overset{CS}{\leq }\sqrt{\sin^2 A +1^2 + \sin^2 A + 2\cos^2 A} \sqrt{\sqrt{2}^2 + 5\sin^2 B +10\cos^2 B + 5\sin^2 B}\\
 &= \sqrt{3}\cdot \sqrt{12} =\sqrt{36}=6
\end{aligned}$$

completing the exercise.

The exercise can also be found in mathematica.gr