Q:

Doubts in proof of topologically transitive + dense periodic points = Devaney Chaotic

Let $(X,d)$ be an infinite metric space with no isolated points and $f : X \to X$ be topologically transitive with dense periodic points. Then $f$ exhibits sensitive dependence on initial data.
The first in the proof given here say that there is $\delta>0$ such that for each $x \in X$ there exists a periodic point $q$ whose orbit $O(q)$ is of distance at least $\frac {\delta}{2}$ from $x.$
We choose two periodic points $q_1,q_2 \in X$ such that $O(q_1) \cap O(q_2) =\emptyset$ and let $\delta = d(O(q_1),O(q_2))>0.$ Then for $x \in X$ and all $i,j$ we have 
$$\delta\leq d(f^i(q_1),f^j(q_2))<d(f^i(q_1),x)+d(x,f^j(q_2))$$
So, either $$d(f^i(q_1),x) \geq \frac {\delta}{2} \text{ or } d(f^j(q_2),x) \geq \frac {\delta}{2}\tag{1}$$
My question:
How does it follow from here that either $$d(O(q_1),x)\geq \frac{\delta}{2} \text{ or } d(O(q_2),x)\geq \frac{\delta}{2}\tag{2}$$
How do we get that for each $i$ and $j$ the same inequality in $(1)$ holds?
Edit:  I think my question lacked clarity. I understand how $(1)$ follows. But I don't understand how $(2)$ follows from $(1)?$ It may happen that when I take $i,j=1,$ then inequality in $(1)$ holds for $q_1$ but when I take different $i,j$ inequality holds for $q_2.$ 

A:

Since the untagged equation is true for all $i,$ therefore
$$\delta\leq d(O(q_1),x)+d(x,f^j(q_2)).$$
And this is true for all $j.$ Thus, 
$$\delta \leq d(O(q_1),x)+d(x,O(q_2)).$$ It follows that 
$$d(O(q_1),x)\geq \frac{\delta}{2} \text{ or } d(O(q_2),x)\geq \frac{\delta}{2}$$

