We have $0\lt a\leq b\leq c\lt \frac{\pi}{2}$, and $A$=($0$, $0$, $1$), $B$=($\sin c$, $0$, $\cos c$) on the unit sphere. Also $$C_\phi=(\sin b\cos \phi, \sin b\sin \phi,\cos b)$$ is given such that $\lvert AC_\phi\rvert = b$ for all $\phi$
Firstly I need to show $$\exists\, \phi\; with\; \lvert BC_\phi\rvert=a \iff a+b\geq c.$$
Then I must deduce the spherical cosine rule from this for a spherical triangle with sides $a,b,c$ and opposite angles $\alpha,\beta,\gamma$ : $$\cos a =\cos b\cos c +\sin b\sin c\cos \alpha.$$ $\space$
For the first part I have gotten as far as showing that $\lvert BC_\phi \rvert=a $ $$\iff \exists\phi \:with\;\cos \phi = \frac{2-2\cos b\cos c - a^2}{2\sin b\sin c}$$ $$\iff\cos (c+b)\leq1- \frac{a^2}{2}\leq\cos (c-b)\,,$$ But I am unsure where to go from here. I'm thinking that $1- \frac{a^2}{2} \leq \cos a$ may be useful since $\cos a \leq \cos(c-b)$ gives the result, but I cannot see how to show this second inequality.
Any guidance would be great. Thank you!