The Inequalities Between Sequences

Suppose there are three sequences {a_n}, {b_n}, and {c_n}, where n=1, 2, 3 … . These sequences are defined recursively, and with inter-dependence; for example, the definition of b_n depends not only on b_n-1, but also on a_n-1 or c_n-1. Suppose we are to work out some inequality relations among a_n, b_n, and c_n. In what way can we make it?

Here is one such question, as follows.

Let p, q be any two distinct positive real numbers such that pq = 5.

Define A_0 = -1 - p - q, B_0 = p+q+pq, C_0 = - pq <0.

For n \geq 1 , define

Show that 1+B_n > (-C_{n-1} +1)^2 >0 for any n \geq 1 .

Let us first calculate B₁. Using the recursive definition, we have that B_1 = (B_0)^2 - 2 A_0 C_0 \\ = (p+q+pq)^2 - 2 (-1-p-q) (-pq) \\ = p^2 + q^2 + p^2 q^2 >=0. Note it is obvious that B₁ is non-negative. Continuing in this way, any B_n shall be non-negative.

Let us set the next immediate goal as 1+B_1 > (-C_0 +1)^2 . Bringing the value of C_0 = -pq , and noting that B_1 = p^2 + q^2 + p^2 q^2 , the goal of proof is equivalent to 1 + p^2 + q^2 + p^2 q^2 > (1-pq)^2; which shall be obvious as p^2 + q^2 > 0 > -2pq .

We list the values of {A₀, B₀, C₀; A₁, B₁, C₁}, in terms of p, q, as follows: (previously, we show how B₁ can be calculated; the read can try to find A₁, C₁, similarly)

A_0 = -1-p-q, B_0 = p+q+pq, C_0 = -pq, \\ A_1 = -1 - p^2 -q^2, B_1 = p^2 + q^2+p^2 q^2, C_1 = -p^2 q^2 .

The quick-minded reader must have found that, when advancing the index by 1, the following schema holds true (where p^*, q^* are two distinct positive real numbers), A_{n-1} = -1-p^*-q^*, C_{n-1} = -p^* q^*, \\B_{n-1} = p^*+q^*+p^*q^*, \\ A_{n} = -1 - (p^*)^2 -(q^*)^2, C_{n} = -{p^*}^2 {q^*}^2, \\ B_{n} = {p^*}^2+{q^*}^2+{p^*}^2 {q^*}^2.

So, a straightforward generalization for the proof of 1+B_1 > (-C_0 +1)^2 , leads to the proof of 1+B_n > (-C_{n-1}+1)^2 .

We have two remarks in sequel.