# Proof by induction that $x_n>2$ where $x_{n+1}=\frac{2x_n^2 +4x_n -2}{2x_n+3}$

The sequence $x_1$ $x_2$ $x_3$..... is such that $x_1=3$ and $$x_{n+1}=\frac{2x_n^2 +4x_n -2}{2x_n+3}$$

Prove by induction that $x_n>2$ for all $n$. First I proved the base case using $n=1$ as $x_1=3>2$. Then assumed it true for $n=k$ where $k$ is some integer. Now I'm trying to prove it for $k+1$. I used the given relationship to get an expression for $x_{k+1}$ but I don't see how I can prove that it's greater than $2$.

• I would phrase the inductive step as follows. Let $k>1$ be an integer and assume the statement true for $n=k$. Then [@Alijah Ahmed answer]. (I like being formal.) – rafforaffo May 28 '15 at 12:20

Rewriting $x_{k+1}$ (and given that $x_k>2$) we have $$x_{k+1}=\frac{2x_k^2 +4x_k -2}{2x_k+3}=x_k+\frac{x_k-2}{2x_k+3}>x_k>2$$

• it should be $\dfrac{2x_k^2 +3x_k + x_k -2}{2x_k+3} = \dfrac{x_k -2}{2x_k+3} + x_k > 2$ ? – Chival May 28 '15 at 10:55
• @Chival: Yes, you are right. Have just corrected the answer. Many thanks! – Alijah Ahmed May 28 '15 at 10:56

Another alternative way to show it is to consider the following.

$f(x)=\frac{2x^2 +4x-2}{2x+3}$

Then you find the derivative with respect to x.

You see for every $x>0$ the derivative is positive therefore the function is increasing.

Therefore $x1<f(x1)=x2<f(f(x1))=x3<...$ and so on.