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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 $t$?

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    @AD. It is from the book "Metric spaces of non-positive curvature" by M.R.Bridson and A. Haefliger, p.20.2011-10-21

3 Answers 3

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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 $\eqalign{z\cosh t+ c\sinh t&=\cosh\tau\cosh t+\sinh\alpha\sinh t\geq \cosh\alpha\cosh t+\sinh\alpha\sinh t\cr &=\cosh(\alpha+t)\geq1\ .\cr}$

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$\sinh t=\frac{1}{2}(e^{t}-e^{-t}) , \cosh t=\frac{1}{2}(e^{t}+e^{-t})$ , 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.

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    @TCL,I'm not familiar with topology so I can only trust you that your proof is correct2011-10-22
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The curve $(X_1,X_2,X_3)=\cosh(t)(x,y,z)+\sinh(t)(a,b,c), -\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 $t$.