Number of positive integers $n$ The question is to find out the number of positive integers $n$ such that $nx^4+4x+3 \leq 0 $ for some real $x$ (without using a graphic calculator).
My attempt at the solution:
We have $nx^4+4x+3 \leq 0 $ which is $nx^4 \leq -3-4x$ or $n \leq -\frac{(3+4x)}{x^4}$. It can easily be observed that for $x=-1$ the condition of $n$ being a positive integer is satisfied. But I could not get others. Moreover to satisfy that $n$ is positive we must have $x<-3/4$.
Please help me in this regard. Thanks.

$x = 0$ cannot be a root so suppose $x > 0$.
The equation is equivalently written as $nx^3+3/x = -4$. 
But by AM GM, $nx^3+3/x \ge 4 \sqrt[4]{n} > -4$ so no positive roots.
Likewise, let $x = -r$ where $r$ is a positive number.
Then we have $nr^4-4r+3 = 0$ or $nr^3+3/r = 4$.
Again by AM GM, we have $nr^3+3/r \ge 4\sqrt[4]{n} \ge 4$ with equality at $n = 1$ 
which is the unique positive integer.
