"The terms of the sequence $x_0,x_1,\ldots$ satisfy


for $n\geq 0$. Prove that is $x_0>3$ then the sequence is strictly increasing."

I am very stuck on dealing with the inequality here. I know I have to use proof by induction but I cannot even figure out how to show that $x_1$ is greater than $x_0$. I was thinking about doing this graphically, but the problem is that when I sketch this as a function, I know that what I am looking for is that $f(x)$ is above $y=x$ for values of $x$ greater than $3$, but the whole graph is actually above or touching $y=x$, which makes no sense to me because surely for values of $x$ less than $3$ the sequence is decreasing, so the function should lie below the $y=x$ line...

Any ideas would be much appreciated!


Hints by induction:

  • $x_1-x_0=\frac{(x_0-3)(x_0-2)}{5}>0$
  • Assume that $x_n>x_{n-1}>\cdots>3$ then $x_{n+1}-x_n=\frac{(x_n-3)(x_n-2)}{5}>0$

All you need to prove is that $a_{n+1} > a_{n}$ for $a_{n}>3$. So it transforms into $a_{n}^{2}-5a_{n}+6>0$, $(a_{n}-3)(a_{n}-2)>0$, which is true, if $a_{n}>3$.


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