A question about hyperbolic functions Suppose $(x,y,z),(a,b,c)$ satisfy $$x^2+y^2-z^2=-1, z\ge 1,$$ $$ax+by-cz=0,$$ $$a^2+b^2-c^2=1.$$ Does it follow that $$z\cosh(t)+c\sinh(t)\ge 1$$ for all real number $t$?
 A: $\sinh t=\frac{1}{2}(e^{t}-e^{-t})  ,  \cosh t=\frac{1}{2}(e^{t}+e^{-t})$ , so we may write:
$\frac{z}{2}(e^{t}+e^{-t})+\frac{c}{2}(e^{t}-e^{-t})\geq 1 \Rightarrow$
$\Rightarrow (\frac{z}{2}+\frac{c}{2})e^t+(\frac{z}{2}-\frac{c}{2})e^{-t}\geq 1$ ,if we make substitution $e^t=p$ we have that:
$(\frac{z}{2}+\frac{c}{2})p^2-p+(\frac{z}{2}-\frac{c}{2})\geq 0$ ,this inequality is true for:
$\frac{z}{2}+\frac{c}{2}>0\Rightarrow z+c>0 \land 1-4(\frac{z^2}{4}-\frac{c^2}{4})\leq 0\Rightarrow (z-c)(z+c)\geq 1$
So we have two conditions:
$z+c>0 \land (z-c)(z+c)\geq 1$
Now if we multiply by two second equation from the text of the question and if we add all three equations we may write following:
$(a+x)^2+(b+y)^2-(z+c)^2=0\Rightarrow$
$\Rightarrow z+c=\pm \sqrt{(a+x)^2+(b+y)^2}$
so if: $c<0 \land|z|<|c|\Rightarrow z+c<0$ which means that first condition isn't satisfied,therefore we may conclude that given inequality from text of the question doesn't follow from these three equalities.
A: The curve  $(X_1,X_2,X_3)=\cosh(t)(x,y,z)+\sinh(t)(a,b,c), -\infty<t<\infty$ is continuous and satisfies $X_1^2+X_2^2-X_3^2=-\cosh^2(t)+\sinh^2(t)=-1$. One of its point $(x,y,z)$ (when $t=0$) lies on the upper sheet $X_1^2+X_2^2-X_3^2=-1, X_3\ge 1$. By connectness of the curve, the whole curve must lie in this connected component. Hence $z\cosh(t)+c\sinh(t)\ge 1$ for all $t$.
A: Write
$$a=\rho\cos\phi,\quad b=\rho\sin\phi;\qquad x=r\cos\psi,\quad y=r\sin\psi.$$
Then $1+r^2=z^2$, whence 
$$r=\sinh \tau,\quad z=\cosh\tau$$
for some $\tau\geq0$. Similarly $c^2=a^2+b^2-1=\rho^2-1$, whence
$$\rho=\cosh\alpha,\quad c=\sinh\alpha$$
for some $\alpha\in{\mathbb R}$. Therefore we get
$$\eqalign{ax+by-cz&=\rho r(\cos\phi\cos\psi+\sin\phi\sin\psi)-\cosh\tau\sinh\alpha \cr
&=\sinh\tau\cosh\alpha\cos(\phi-\psi)-\cosh\tau\sinh\alpha\ .\cr}$$
As $ax+by-cz=0$ this implies
$$\left|{\sinh\alpha\over\cosh\alpha}\right|\leq\left|{\sinh\tau\over\cosh\tau}\right|$$
or $|\alpha|\leq\tau$.
It follows that for any real $t$ one has
$$\eqalign{z\cosh t+ c\sinh t&=\cosh\tau\cosh t+\sinh\alpha\sinh t\geq \cosh\alpha\cosh t+\sinh\alpha\sinh t\cr &=\cosh(\alpha+t)\geq1\ .\cr}$$
