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.
 A: In your approach, choosing $n=1$ works for the first inequality, but is too restrictive for the second inequality, because $\frac{5}{5x^2 - 5x +2}$ becomes smaller than $4/n=4$ when $x$ is large.
Note that the quadratic polynomials at the denominators are positive. It follows that the given double-inequality is equivalent to
$$\frac{4(5x^2 - 5x +2)}{5} < n < \frac{4(3x^2 - 3x + 2)}{3}.$$
Since $n$ should be an integer (it has to be positive by the above inequality), ad any interval whose size is greater than $1$ contains an integer, it  suffices to show that
$$\frac{4(3x^2 - 3x + 2)}{3}-\frac{4(5x^2 - 5x +2)}{5}>1$$
which is true because the left-hand side is equal to $16/15$.
A: Hint: write the inequalities as$\frac {4(5x^{2}-5x+2)} 5 <n <\frac 4 3 (3x^{2}-3x+2)$. To show that there is a positive integer satisfying this it is enough to very two facts:
a) $\frac 4 3 (3x^{2}-3x+2) -\frac {4(5x^{2}-5x+2)} 5 >1$
and 
b)  $\frac {4(5x^{2}-5x+2)} 5 >0$.
To prove a) simplify the quadratic expression and complete the square
(The quadratic expressions in the question are always positive). 
A: 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?
