# Angles of a triangle

• Sep 10th 2009, 08:23 PM
dhiab
Angles of a triangle
Prove that if $\displaystyle A,B,C$ are angles of a triangle , then :
$\displaystyle \begin{array}{l} 1)\cos A + cosB + \cos C \le \frac{3}{2} \\ 2)\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \le \frac{1}{8} \\ \end{array}$
• Sep 11th 2009, 10:12 AM
red_dog
1) $\displaystyle 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}+\cos C\leq\frac{3}{2}$

$\displaystyle 2\sin\frac{C}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}\leq\frac{3}{2}$

$\displaystyle 2\sin^2\frac{C}{2}-2\cos\frac{A-B}{2}\sin\frac{C}{2}+\frac{1}{2}\geq 0$

Let $\displaystyle f(x)=2x^2-2\cos\frac{A-B}{2}x+\frac{1}{2}$

The discriminant is $\displaystyle \Delta=4\left(\cos^2\frac{A-B}{2}-1\right)\leq 0\Rightarrow f(x)\geq 0, \ \forall x\in\mathbb{R}$

2) Is the same inequality because

$\displaystyle \cos A+\cos B+\cos C=4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}+1$
• Sep 11th 2009, 02:47 PM
DavidEriksson
A shorter solution for 1. It's enough to show that it's true for $\displaystyle A,B,C\in[0,\frac{\pi}{2}]$.

Now to the proof for this case. Notice that $\displaystyle \cos x$ is a concave on $\displaystyle [0,\frac{\pi}{2}]$ so by Jensens inequality:
$\displaystyle \frac{\cos A + \cos B + \cos C}{3} \leq \cos \left (\frac{A+B+C}{3} \right )=\cos \frac{\pi}{3} = \frac{1}{2}$ with equality iff $\displaystyle A=B=C=\frac{\pi}{3}$.