Is there any solution to find a condition for $f(x)=a+bx^n+cx^2-dx>0$ to always hold true? Okay, I am interested to know the criteria for a function to always hold $$f(x)=a+bx^n+cx^2-dx>0,$$
if it is given that $a, b, c>0$ and $n\in(-2,2)$ is some real number and $x>0$.  My idea was to find a minima of this function, and at minima $x_m$, the following condition has to be satisfied $$nbx_m^{n-1}+2cx_m-d=0,$$ and the function exhibits minima only when $$n(n-1)x_m^{n-2}>-2c.$$
Now to find a condition when $f(x)>0$ is always satisfied, my idea would be to find $x_m$ from second equation and replace it in the first equation. The problem is second equation is not solvable except for some special values of $n$. So my question is it this the dead end or there is something more that can be done to approach towards the solution? Any inputs would be greatly appreciated.
 A: Take, for example, the case  $f(x) = x^6+x^2-x + a$.  Here $f$ is convex, with a minimum at the real root $r$ of $f'(x) = 6 x^5 + 2 x - 1$, which happens to be approximately $.4466635719$.  The condition to have  $f(x) > 0$ for all $x>0$ is $a > -r^6 - r^2 + r$.  It can be shown that $-r^6 - r^2 + r$ is a root of the polynomial $  -{z}^{5}-{\frac {8\,{z}^{3}}{27}}+{\frac {25\,{z}^{2}}{27}}-{\frac {
5881\,z}{11664}}+{\frac{1127}{15552}}
$, which has Galois group $S_5$ and is not solvable by radicals.  So unless you can solve quintic equations, you won't be find an explicit necessary and sufficient condition for your problem.  Of course you can find approximate solutions, e.g. by using numerical methods to approximate the roots of $f'(x)$.
A: Trivial case: $d\le 0$, the inequality is always satisfied. 
Therefore we assume always $d>0$.
The general case independent of $n$ we get with the maximum of $h(x):=xd-cx^2$ :
$h^\prime(x)=0$ => maximum of $h(x)$ is $h(\frac{d}{2}) =\frac{d^2}{4c}$
The inequality above is always fulfilled with $a+bx^n>\frac{d^2}{4c}$. 
Therefore: Should the inequality valid for all $x>0$, then it has to be $d^2\le 4ac$.
Now the cases depending on $n$ because of $d^2>4ac$.
$n=0$ or $n=1$: quadratic equation  (not a problem to discuss)
Therefore: $n\in (-2,+2)\setminus\{0,1\}$
A note at the beginning:
$a+x(bx^{n-1}+cx)>dx$ means, that the inequality is always fulfilled with $bx^{n-1}+cx\ge d$, therefore the cases $0<bx^{n-1}+cx<d$ and $bx^{n-1}+cx\ge d$ have to be discussed.
$n>1$: It doesn’t exist any extremum for $bx^{n-1}+cx$, no criteria can be given independent of $x$.
$n<1$: We get no maximum, only a minimum. => No criteria for $0<bx^{n-1}+cx<d$, but interesting for the other case.
$(bx^{n-1}+cx)^\prime=(n-1)bx^{n-2}+c:=0$ => $x=(\frac{-c}{b(n-1)})^{\frac{1}{n-2}}$ with $\frac{-c}{b(n-1)}>0$
We put the result into $bx^{n-1}+cx\ge d$ and therefore the criteria for the inequality is
$c\frac{n-2}{n-1}(\frac{-c}{b(n-1)})^{\frac{1}{n-2}}\ge d$.
So, I hope that I have calculated right and it helps.
EDIT: Calculation corrected.
EDIT 2: The case $c\to 0^+$ means $c:=0$ and therefore $f(x)=a+bx^n-dx$. 
Extremum (here: minimum): $f^\prime(x)=nbx^{n-1}-d:=0$ => $x=(\frac{d}{nb})^\frac{1}{n-1}$.
This can only exist for $n>0$. We put this result into the inequality and get the condition $a>d\frac{n-1}{n}(\frac{d}{nb})^\frac{1}{n-1}$. For the special case $0<n<1$ is $f(x)>0$ valid independent of $b$ and $d$.
For $n<0$ we always have a solution for $f(x)=0$. This means with $f(x_0)=0$, that we get $f(x_0)>0$ only for $x<x_0$.   
