For all $x$ in $\mathbb{R}$, there exists $n$ such that the below inequality holds (Proof) 
Prove that for all $x \in \mathbb{R}$, there exists $n \in \mathbb{N}$ such that:

$$ \frac{3}{3x^2 - 3x + 2} < \frac{4}{n} < \frac{5}{5x^2 - 5x +2} .$$

Here's how I went about it:
For the first part, I reasoned that if we choose $n = 1$ we satisfy the equation since $\frac{3}{3x^2 - 3x + 2}$ attains a maximum of $\frac{12}{5} < 4$, and us bounded below by zero. So, we can write:
$$3x^2 - 3x + 2 > n \implies \frac{1}{3x^2 - 3x + 2} < \frac{1}{n} \implies\frac{3}{3x^2 - 3x + 2} < \frac{3}{n} < \frac{4}{n} $$
For the second part however, I am having some difficulty. I've tried the same approach as above and have not had success.
Disclaimer: I have not done any sort of math in quite a few years, and therefore feel that I am (a) not sure that my approach is even valid (b) feel that I am missing something incredibly obvious. 
Any help is greatly appreciated.

Since $5x^2-5x+2>0$ and $3x^2-3x+2>0$, we have $$\frac{4}{5}(5x^2-5x+2)<n<\frac{4}{3}(3x^2-3x+2).$$ Now, we see that it's enough to prove that:
$$\frac{4}{3}(3x^2-3x+2)-\frac{4}{5}(5x^2-5x+2)>1.$$
Can you end it now?
