Does Law of Cosines work?

Let's use the identity that $cos(\alpha) + cos(\beta) = 2cos\left(\dfrac{\alpha+\beta}{2}\right)cos\left(\dfrac{\alpha-\beta}{2}\right)$ .

Further, we can rewrite $cos(\gamma) = cos(\pi - \alpha - \beta) = -cos(\alpha + \beta) = -2cos^2\left(\dfrac{\alpha+\beta}{2}\right)+1$

Now let $x = cos\left(\dfrac{\alpha+\beta}{2}\right)$ and $y = cos\left(\dfrac{\alpha-\beta}{2}\right)$ .

Then the expression that we are trying to maximize becomes $2xy - 2x^2 + 1$ , such that $-1 \le x,y \le 1$ .

This new function can be rewritten as $-2\left(x-\dfrac{y}{2}\right)^2+1+\dfrac{y^2}{2}$

Now $-2\left(x-\dfrac{y}{2}\right)^2 \le 0$ and $\dfrac{y^2}{2} \le \dfrac{1}{2}$ .

Hence $2xy - 2x^2 + 1 \le \boxed{\dfrac{3}{2}}$

Note that this value is achieved when $\alpha = \beta = \gamma = \dfrac{\pi}{3}$ .

Using trigonometric identities:

$\cos\alpha+\cos\beta+\cos\gamma=$

$=2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)-\cos\left(\alpha+\beta\right)$

$=2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)+1-2\cos^2\left(\frac{\alpha+\beta}{2}\right)$

$=1+2\cos\left(\frac{\alpha+\beta}{2}\right)\left[\cos\left(\frac{\alpha-\beta}{2}\right)-\cos\left(\frac{\alpha+\beta}{2}\right)\right]$

$=1+4\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}.$

We know that $\frac{\alpha}{2}$ , $\frac{\beta}{2}$ , and $\frac{\gamma}{2}$ are acute angles, for $\alpha+\beta+\gamma=\pi$ . Thus, $\sin\frac{\alpha}{2}$ , $\sin\frac{\beta}{2}$ , and $\sin\frac{\gamma}{2}$ are nonnegative real numbers, and, by the AM-GM inequality, the maximum of $\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}$ holds if and only if $\sin\frac{\alpha}{2}=\sin\frac{\beta}{2}=\sin\frac{\gamma}{2}$ , or, better, $\alpha=\beta=\gamma$ .

Hence, the maximum value of $\cos\alpha+\cos\beta+\cos\gamma$ is $3\cos\frac{\pi}{3}=\boxed{\frac{3}{2}}$ .