Let $(x_{n})_{n\geq 1}$ be a sequence, defined by $x_{1}=6, x_{2}=7$ and $$ x_{n+2}=\frac{4x_{n+1}^{3}-x_{n+1}^{2}x_{n}}{x_{n+1}x_{n} +x_{n}^{2}} $$ Prove that this sequence is strictly increasing. I know that I have to prove that $x_{n+2}-x_{n+1}>0$, but I don't know how.
My attempt:
$$x_{n+2}-x_{n+1}=\frac{4x_{n+1}^{3}-x_{n+1}^{2}x_{n}}{x_{n+1}x_{n} +x_{n}^{2}}-\frac{4x_{n}^{3}-x_{n}^{2}x_{n-1}}{x_{n}x_{n-1} +x_{n-1}^{2}} $$ And now I need to show that this is $>0$.
When you evaluate $x_{n+2}-x_{n+1}$, you don't have to substitute $x_{n+1}$, we can get a simpler formula just by substituting $x_{n+2}$.
For an induction proof, it's clear that $x_2 > x_1$. Now, assume that for some fixed $n\geq 2$, we have $x_{n+1}>x_n$, and we will prove that $x_{n+2}>x_{n+1}$, that is $x_{n+2}-x_{n+1}>0$. Notice that, using $x_{n+1}-x_n>0$, we have:
$$ \begin{aligned} x_{n+2}-x_{n+1}&=\frac{4x_{n+1}^{3}-x_{n+1}^{2}x_{n}}{x_{n+1}x_{n} +x_{n}^{2}}-x_{n+1}\\ &=\frac{4x_{n+1}^{3}-2x_{n+1}^{2}x_{n}-x_n^2x_{n+1}}{x_{n+1}x_{n}+x_n^2}\\ &=\frac{x_{n+1}^3+2x_{n+1}^2(x_{n+1}-x_n)+x_{n+1}(x_{n+1}^2-x_n^2)}{x_{n+1}x_{n}+x_n^2} \end{aligned} $$
and this is stricly positive, because $x_{n+1}-x_n>0$.