Let $\alpha$, $\beta$, and $\gamma$ be acute angles such that $\alpha$ + $\beta$ = $\gamma$. Show that Let $\alpha$, $\beta$, and $\gamma$ be acute angles such that $\alpha$ + $\beta$ = $\gamma$. Show that
cos$\alpha$ + cos$\beta$ + cos$\gamma$ - 1 $\geq$ $2\sqrt{\cos\alpha\times\cos\beta\times\cos\gamma}$.
Here's what I've tried. I know that $0 < \alpha, \beta, \gamma < \pi/2$ since they are acute angles. So I substituted $\alpha+\beta$ for $\gamma$ 
$\cos\alpha$ + $\cos\beta$ + $\cos(\alpha+\beta)$- 1 $\geq$ $2\sqrt{\cos\alpha\cos\beta\cos(\alpha+\beta)}$.
$\cos\alpha + \cos\beta + \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)- 1 \geq 2\sqrt{\frac{1}{2}[\cos(\alpha-\beta)+\cos(\alpha+\beta)]\cos(\alpha+\beta)}.$
$\cos\alpha + \cos\beta + \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)- 1 \geq 2\sqrt{\frac{1}{2}[\cos(\alpha-\beta)\cos(\alpha+\beta)+\cos^{2}(\alpha+\beta)]}$
$\cos\alpha + \cos\beta + \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)- 1 \geq 2\sqrt{\frac{1}{4}[\cos(-2\beta)+\cos(2\alpha)]+\cos^{2}(\alpha+\beta)}$
I got stuck from there and I'm not sure where to go from here. Any help would be appreciated.
 A: Let $\alpha\geq\beta$ 
Since $\cos$ is a decreasing function on $\left(0,\frac{\pi}{2}\right)$, $\frac{\alpha+\beta}{2}<\frac{\pi}{4}$ and $\frac{\alpha-\beta}{2}<\frac{\pi}{4}$, we obtain: 
$$\cos\alpha+\cos\beta+\cos(\alpha+\beta)-1=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}+2\cos^2\frac{\alpha+\beta}{2}-2>$$
$$>2\cdot\frac{1}{\sqrt2}\cdot\frac{1}{\sqrt2}+2\cdot\left(\frac{1}{\sqrt2}\right)^2-2=0$$
Let $\cos\alpha=x$, $\cos\beta=y$ and $\cos(\alpha+\beta)=z$.
Hence, we need to prove that $$(x+y-1+z)^2\geq4xyz$$ or
$$z^2+2(x+y-1-2xy)z+(x+y-1)^2\geq0.$$ 
$$\frac{\Delta_{z}}{4}=(x+y-1-2xy)^2-(x+y-1)^2=$$
$$=(x+y-1-2xy-(x+y-1))(x+y-1-2xy+x+y-1)=$$
$$=4xy(1-x)(1-y).$$
Thus, it remains to prove that
$$z\leq-(x+y-1-2xy)-\sqrt{4xy(1-x)(1-y)}$$ or
$$z\leq(1-x)(1-y)+xy-\sqrt{4xy(1-x)(1-y)}$$ or
$$\cos(\alpha+\beta)\leq\cos\alpha\cos\beta+(1-\cos\alpha)(1-\cos\beta)-2\sqrt{\cos\alpha\cos\beta(1-\cos\alpha)(1-\cos\beta)}$$ or
$$-\sin\alpha\sin\beta\leq4\sin^2\frac{\alpha}{2}\sin^2\frac{\beta}{2}-4\sin\frac{\alpha}{2}\sin\frac{\beta}{2}\sqrt{\cos\alpha\cos\beta}$$ or
$$-\cos\frac{\alpha}{2}\cos\frac{\beta}{2}\leq\sin\frac{\alpha}{2}\sin\frac{\beta}{2}-\sqrt{\cos\alpha\cos\beta}$$ or
$$\cos\frac{\alpha-\beta}{2}\geq\sqrt{\cos\alpha\cos\beta}$$ or
$$1+\cos(\alpha-\beta)\geq2\cos\alpha\cos\beta$$ or
$$1\geq\cos(\alpha+\beta)$$
Done!
A: Using that
$$
\cos\alpha\cos\beta-\cos\gamma = \sin\alpha\sin\beta
 = 4 \sin\frac\alpha2 \cos\frac\alpha2 \sin\frac\beta2 \cos\frac\beta2 \ge 0
$$
and
$$
(1-\cos\alpha)(1-\cos\beta) = 4 \sin^2\frac\alpha2 \sin^2\frac\beta2,
$$
a possible proof:
$$
\cos\gamma - 2\sqrt{\cos\alpha\cos\beta}\cdot\sqrt{\cos\gamma}+
\cos\alpha\cos\beta
\stackrel{?}{\ge} 
(1-\cos\alpha)(1-\cos\beta) 
$$
$$
\Big(\underbrace{\sqrt{\cos\alpha\cos\beta}-\sqrt{\cos\gamma}}_{\ge0}\Big)^2
\stackrel{?}{\ge} 
4\sin^2\frac\alpha2 \sin^2\frac\beta2
$$
$$
\sqrt{\cos\alpha\cos\beta}-\sqrt{\cos\gamma}
\stackrel{?}{\ge} 
2 \sin\frac\alpha2 \sin\frac\beta2
$$
$$
\cos\alpha\cos\beta 
\stackrel{?}{\ge} 
\Big(\sqrt{\cos\gamma} + 2 \sin\frac\alpha2 \sin\frac\beta2\Big)^2
$$
$$
\cos\alpha\cos\beta-\cos\gamma
\stackrel{?}{\ge} 
4 \sin\frac\alpha2 \sin\frac\beta2 \sqrt{\cos\gamma} 
+ 4 \sin^2\frac\alpha2 \sin^2\frac\beta2
$$
$$
4 \sin\frac\alpha2 \cos\frac\alpha2 \sin\frac\beta2 \cos\frac\beta2 \stackrel{?}{\ge} 
4 \sin\frac\alpha2 \sin\frac\beta2 \sqrt{\cos\gamma} 
+ 4 \sin^2\frac\alpha2 \sin^2\frac\beta2
$$
$$
 \cos\frac\alpha2 \cos\frac\beta2
-  \sin\frac\alpha2 \sin\frac\beta2
\stackrel{?}{\ge} 
\sqrt{\cos\gamma} 
$$
$$
\cos\frac\gamma2 \stackrel{?}{\ge} 
\sqrt{\cos\gamma} 
$$
$$
\sqrt{\frac{1+\cos\gamma}2} \stackrel{?}{\ge} 
\sqrt{\cos\gamma} 
$$
$$
1 \ge \cos\gamma.
$$
