As a part of an induction proof, I need to prove that if $\frac{3x+2}{x+2} < x$, then $\frac{3x+2}{x+2} > \frac{11x+10}{5x+6}$. I need to prove this for $x>0$ if that matters so basically $x$ has to be greater than 2. I have tried manipulating this inequality in a number of ways but it always leads to a dead end. How can I prove this?
Edit:
Managed to prove this like following:
From the assumption we get that $x>2$. So considering this $$\frac{11x+10}{5x+6} < \frac{11x+10}{4x+8} < \frac{12x+8}{4x+8} = \frac{3x+2}{x+2}$$
Hint: For $$x>0$$ we get that $$\frac{3x+2}{x+2}<x$$ is equivalent to $$0<x^2-x-2$$ and this is $$0<(x+1)(x-2)$$ so it must be $$x>2$$ and now we will Show that $$\frac{3x+2}{x+2}>\frac{11x+10}{5x+6}$$ is true for $$x>2$$ this is $$(3x+2)(5x+6)>(11x+10)(x+2)$$ Expanding and collecting like terms we get $$4x^2-4x-8>0$$ dividing by $4$ we get $$x^2-x-2>0$$ and this is $$(x+1)(x-2)>0$$ true for $$x>2$$!!!