# How to prove inequalities between $U_{n+1}$ and $U_n$?

I meet a lot of questions in multiple series exercise that asks me to prove some inequalities between $U_{n+1}$ and $U_n$ in general .

I'm going to introduce 2 sample questions I've met recently . I'm looking into how to approach such problems instead of how to solve those samples

Exemple:

Consider the function $f(x) = \frac{4x}{x + 2}$ and the serie $U_n$ defined by recurency where $U_0 = 1$ and $U_{n+1} = f(U_n)$

Given thatI've already proved that U_n is convergent toward 2 and increasing and that $1<U_n<2$, I need to prove the following :

$$|U_{n+1} - 2| <= \frac{2}{3}|U_n - 2|$$

Another Example:

Given the following serie : $W_0 = 1$ and $W_{n+1} = \sqrt{5W_n + 6}$ and that I've already demonstrated that $W_n$ is increasing and that $1< W_n < 6$ , I need to prove that :

$$6 - W_{n+1} <= \frac{5}{6}(6 - W_n)$$

As you can see, in bith example I need to prove some kind of inequality between $U_{n+1}$ and $U_n$, even though that the inequality is different, I believe they both share the same idea or principle, Am I wrong ?

So How can I approach solving this kind of inequalities in general ? Should I use recursive proofs ? Or maybe the first inequalities I've proved ( $a < U_n < b$) ? I'm always stuck at these kind of questions .

• If I were you I would do a little bit about the behavior of functions in the neighborhoods of its fixed points. – Franklin Pezzuti Dyer May 23 '17 at 22:35
• We didn't approach this in class, so I don't think I can use this, I'm looking more into algebraic approach and simple inequalities manipulation techniques . – Anis Souames May 23 '17 at 22:37
• Okay, I posted an answer using an algebraic approach. – Franklin Pezzuti Dyer May 23 '17 at 22:50

First of all, since you have proven that $1\lt U_n \lt 2$, we know that $$U_n-2$$ will always be negative, and so its absolute value will be $$2-U_n$$ Now we can change your inequality to $$2-U_{n+1}\le\frac{4}{3}-\frac{2}{3}U_n$$ Just as a tip, if I were you, I would always try to get rid of the absolute value first thing. Inequalities with absolute value are unpleasant to deal with.
You could prove this using simple substitution. When we substitute, we end up with $$2-\frac{4U_n}{U_n+2} \le \frac{4}{3}-\frac{2}{3}U_n$$ $$6-\frac{12U_n}{U_n+2} \le 4-2U_n$$ $$6-\frac{12U_n}{U_n+2} \le -2(U_n-2)$$ Since we know that $U_n+2$ is positive always, we can multiply both sides by it: $$6-12U_n \le -2(U_n^2-4)$$ $$0 \le -2U_n^2+12U_n-10$$ $$0 \ge U_n^2-6U_n-5$$ Since we know that $1\lt U_n \lt 2$, we can see that this must always be true because $U_n^2$ cannot equal or exceed $2U_n$.
The inequality you found bounding $U_n$ was definitely helpful... that's a good thing to look for when doing inequalities of this type.