# Thread: Angles of a triangle

1. ## Angles of a triangle

Prove that if $A,B,C$ are angles of a triangle , then :
$

\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}

$

2. 1) $2\cos\frac{A+B}{2}\cos\frac{A-B}{2}+\cos C\leq\frac{3}{2}$

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

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

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

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

2) Is the same inequality because

$\cos A+\cos B+\cos C=4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}+1$

3. A shorter solution for 1. It's enough to show that it's true for $A,B,C\in[0,\frac{\pi}{2}]$.

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