$H(\xi_2) \geq H(\xi)^{\kappa-1}$ Let $\xi=\dfrac{x}{y} \in \mathbb{Q}$ and define height function $H(\xi)=\max(|x|,|y|)$.

Let $\alpha$ be a real number, and $\kappa>2$, and consider the inequality
  $$\left| \xi-\alpha \right| \leq H(\xi)^{-\kappa} \; \text{in} \;\xi \in \mathbb{Q} \; \text{with} \; \xi >\alpha$$
  Prove that if $\xi_1,\xi_2$ are two distinct solutions of this inequality with $H(\xi_2) \geq H(\xi_1)$, then 
  $$H(\xi_2) \geq H(\xi_1)^{\kappa-1}$$

Let $\xi_1=\dfrac{a_1}{b_1}$ and $\xi_2=\dfrac{a_2}{b_2}$ then 
$$\left| \xi_1 - \xi_2 \right|= \left| \dfrac{a_1}{b_1} - \dfrac{a_2}{b_2} \right|=\left| \dfrac{a_1b_2-a_2b_1}{b_1b_2} \right|$$
Since $a_1,a_2,b_1,b_2 \in \mathbb{Z}$ and $\xi_1,\xi_2$ are distinct, $0\neq a_1b_2-a_2b_1 \in \mathbb{Z}$. Also, $|b_1| \leq H(\xi_1)$ and $|b_2| \leq H(\xi_2)$ hence
$$  \left| \xi_1 - \xi_2 \right| \geq \left(H(\xi_1)H(\xi_2)\right)^{-1}$$
Therefore
$$\left(H(\xi_1)H(\xi_2)\right)^{-1} \leq \left| \xi_1 - \xi_2 \right| \leq |\xi_1-\alpha| + |\xi_2 -\alpha| \leq H(\xi_1)^{-\kappa}+H(\xi_2)^{-\kappa} \leq 2 H(\xi_1)^{-\kappa}$$
So I get $2H(\xi_2) \geq H(\xi_1)^{\kappa-1}$ but not $H(\xi_2) \geq H(\xi_1)^{\kappa-1}$. 
Could anyone suggest me stronger estimate? I think we should exploit condition $\xi >\alpha$ but doesn't know how to do.
 A: First, the question is not currently well-posed in that the $H$ function does not give a unique value. I assume you also want $\xi=\dfrac{x}{y} \in \mathbb{Q}$ to be in lowest terms, i.e., so $\gcd(x,y) = 1$.
You have the right idea, but there's a more useful way to deal with $\left| \xi_1 - \xi_2 \right|$. You're correct you should explicitly exploit the condition $\xi \gt \alpha$. In particular, you get
$$\xi_1 = \alpha + c_1, \; c_1 \gt 0 \tag{1}\label{eq1A}$$
$$\xi_2 = \alpha + c_2, \; c_2 \gt 0 \tag{2}\label{eq2A}$$
Thus, you then have
$$\begin{equation}\begin{aligned}
\left| \xi_1 - \xi_2 \right| & = \left|(\alpha + c_1) - (\alpha + c_2) \right| \\
& = \left|c_1 - c_2 \right| \\
& \lt \max(c_1, c_2)
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
If $\max(c_1, c_2) = c_1 = \xi_1 - \alpha$, you then get
$$\begin{equation}\begin{aligned}
\left(H(\xi_1)H(\xi_2)\right)^{-1} & \leq \left| \xi_1 - \xi_2 \right| \\
& \lt \left| \xi_1 - \alpha \right| \\
& \le H(\xi_1)^{-\kappa}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Multiplying both sides by $H(\xi_1)^{\kappa}H(\xi_2)$ gives
$$H(\xi_2) \gt H(\xi_1)^{\kappa - 1} \tag{5}\label{eq5A}$$
Consider instead that $\max(c_1, c_2) = c_2 = \xi_2 - \alpha$, you then get
$$\begin{equation}\begin{aligned}
\left(H(\xi_1)H(\xi_2)\right)^{-1} & \leq \left| \xi_1 - \xi_2 \right| \\
& \lt \left| \xi_2 - \alpha \right| \\
& \le H(\xi_2)^{-\kappa}
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
Multiplying both sides by $H(\xi_2)^{\kappa}H(\xi_1)$ gives
$$H(\xi_1) \gt H(\xi_2)^{\kappa - 1} \tag{7}\label{eq7A}$$
However, from the given condition of $H(\xi_2) \geq H(\xi_1)$, since the values are positive, $k \gt 2$ and $H(\xi_1) \ge 1$, you have
$$H(\xi_2)^{\kappa-1} \geq H(\xi_1)^{\kappa-1} \ge H(\xi_1) \tag{8}\label{eq8A}$$
This, which combined with \eqref{eq7A}, gives
$$H(\xi_1) \gt H(\xi_2)^{\kappa - 1} \ge H(\xi_1)^{\kappa - 1} \ge H(\xi_1) \implies H(\xi_1) \gt H(\xi_1) \tag{9}\label{eq9A}$$
This, of course, is not possible. Thus, the only available possibility is given in \eqref{eq7A}, which is even slightly stronger than what you were asked to prove.
A: Here is my own answer that can take advantage of condition $\xi >\alpha$. 
Case 1: $|a_1b_2-a_2b_1| \geq 2$, we have:
$$2\left(H(\xi_1)H(\xi_2)\right)^{-1} \leq \left| \xi_1 - \xi_2 \right| \leq |\xi_1-\alpha| + |\xi_2 -\alpha| \leq H(\xi_1)^{-\kappa}+H(\xi_2)^{-\kappa} \leq 2 H(\xi_1)^{-\kappa}$$
Thus $H(\xi_2) \geq H(\xi_1)^{\kappa-1}$
Case 2: $|a_1b_2-a_2b_1|=1$. If $a_1b_2=a_2b_1+1$ then $\dfrac{a_2}{b_2}=\dfrac{a_1}{b_1}-\dfrac{1}{b_1b_2}$. We have $\xi_2 \geq \alpha$, so
$$0 \leq \dfrac{a_2}{b_2}-\alpha=\dfrac{a_1}{b_1}-\dfrac{1}{b_1b_2}-\alpha$$
Therefore $(H(\xi_1)H(\xi_2))^{-1} \leq \dfrac{1}{b_1b_1} \leq \dfrac{a_1}{b_1} -\alpha \leq H(\xi_1)^{-\kappa}$ and $H(\xi_2) \geq H(\xi_1)^{\kappa-1}$.
If $a_1b_2+1=a_2b_1$ then $\dfrac{a_1}{b_1}=\dfrac{a_2}{b_2}-\dfrac{1}{b_1b_2}$. We have $\dfrac{a_1}{b_1}-\alpha \geq 0$ so 
$$H(\xi_1)^{-\kappa} \geq H(\xi_2)^{-\kappa} \geq \dfrac{a_2}{b_2}-\alpha \geq \dfrac{1}{b_1b_2} \geq (H(\xi_1)H(\xi_2))^{-1}$$
We also get $H(\xi_2) \geq H(\xi_1)^{\kappa-1}$.
