# An inequality with Cosines and a Sine

### Solution 1

On the interval $\displaystyle \left[0,\frac{\pi}{2}\right],\,$ function $\cos x\,$ is concave, such that, by Jensen's inequality,

$\displaystyle \cos A+\cos B\le 2\cos\frac{A+B}{2}=2\sin{C}{2}.$

Hence suffice it to prove that $\displaystyle \cos B+2\cos\frac{A+B}{2}\le 3\cos\frac{\pi+B-C}{3}.\,$ We again apply Jensen's inequality:

\displaystyle \begin{align} \cos B+2\cos\frac{A+B}{2} &=\cos A+\cos\frac{A+B}{2}+\cos\frac{A+B}{2}\\ &\le 3\cos \left(\frac{\displaystyle B+\frac{A+B}{2}+\frac{A+B}{2}}{3}\right)\\ &=3\cos\frac{A+2B}{3} \end{align}

and note that, since $A+B+C=\pi,\,$ $\displaystyle \frac{A+2B}{3}=\frac{\pi+B-C}{3}.$

### Solution 2

Observe that $\displaystyle \sin\alpha=\cos\left(\frac{\pi}{2}-\alpha\right)\,$ and make use of Jensen's inequality as above:

\displaystyle \begin{align} \cos A+4\cos B+4\sin\frac{C}{2} &= \cos A+4\cos B+4\cos\left(\frac{\pi}{2}-\frac{C}{2}\right)\\ &\le 9\cos\left(\frac{\displaystyle A+4B+4\left(\frac{\pi}{2}-\frac{C}{2}\right)}{3}\right)\\ &=9\cos\frac{2\pi+A+4B-2C}{9}\\ &=9\cos\frac{3\pi+3B-3C}{9}=9\cos\frac{\pi+B-C}{3}. \end{align}

### Acknowledgment

The problem (from the Romanian Mathematical Magazine) has been posted by Dan Sitaru at the CutTheKnotMath facebook page, and commented on by Leo Giugiuc with his (Solution 1). Solution 2 may seem as a slight modification of Solution 1.