In this section, we shall prove our theorem directly. First we consider the case that is an even number. It is clear that in this case our theorem is equivalent to
or
(2)
Now we prove that for all positive integers k, we have the inequality
(3)
It is clear that (3) holds for . So, without loss of generality, we can assume that . Note that , , , and
so inequality (3) is equivalent to
(4)
From the definition and properties of and , we can easily deduce the identities
So, applying these formulae, we have
and
From these two identities and (4), we deduce that inequality (3) is equivalent to
(5)
For convenience, we let
and
Then by calculation it follows that
Observe that the major terms of A and B (above those of order ) are in total agreement. Note that , we have
for all integers . So, inequalities (3), (4) and (5) hold for all integers .
Now, applying (3) repeatedly, we have
(6)
On the other hand, we prove the inequality
(7)
This inequality is equivalent to
or
or
(8)
It is clear that inequality (8) holds for all integers , so inequality (7) is true. Now, applying (7) repeatedly, we have
(9)
Combining (6) and (9), we may immediately deduce inequality (2).
Now we consider that is an odd number. It is clear that in this case our theorem is equivalent to
or
(10)
First we prove the inequality
(11)
It is easy to check that inequality (11) is correct for . So, we can assume that . Note that , , , , , so inequality (11) is equivalent to the inequality
(12)
From the definition and properties of the Pell-Lucas numbers, we have
and
By these two identities and (12), we deduce that inequality (11) is equivalent to
(13)
For convenience, we let
and . Then we have
Note that , we have
for all integers . So, inequalities (11), (12) and (13) hold for all integers .
Now, applying (11) repeatedly, we have
(14)
On the other hand, we prove the inequality
(15)
It is easy to check that inequality (15) is correct for . So, we can assume that . This time, inequality (15) is equivalent to
or
or
(16)
It is clear that inequality (16) holds for all integers , so inequality (15) is true. Now, applying (15) repeatedly, we have
(17)
Combining (14) and (17), we may immediately deduce inequality (10).
Now our theorem follows from inequalities (2) and (10). This completes the proof of our theorem.