Find all $n\in\mathbb{Z^+}$ such that $3x^n+n(x+2)-3\ge nx^2$ for all $x\in\mathbb{R}$ 
Find all $n\in\mathbb{Z^+}$ such that $3x^n+n(x+2)-3\ge nx^2$ for all $x\in\mathbb{R}$

This question to me is tricky and I don't know where to start. I tried to substitute $n$ with multiple values like $1$ but I couldn't find a solution. For case $n=1$ I get that $4x-1\ge x^2$ which is not true for all $x$. The fact that $x$ can be any real value is troubling for me. I also tried to use induction to prove that some cases are incorrect but couldn't due to the fact that $x$ isn't a fixed value. Any help would be appreciated. 
 A: The solutions to your problem are the even values of $n$.
Let $f_n(x)=3x^n+n(x+2)-3-nx^2$.
If $n$ is odd, then either $n=1$ in which case the leading monomial in $f_n(x)$
is $-x^2$, or $n\geq 3$ in which case the leading monomial in $f_n(x)$
is $3x^n$. In both cases, $f(x)\to -\infty$ when $x\to -\infty$ so $f$ cannot
be nonnegative.
So suppose now that $n$ is even. We will show that $f_n(x) \geq 0$.
For $n=2$, $f_2(x)=(x+1)^2 \geq 0$. For $n=4$, $f_4(x)=((x+1)^2)(3x^2-6x+5)\geq 0$. So we may assume $n\geq 6$.
If $x\leq -1$, we can write $x=-1-y$ where $y\geq 0$, and then
$$
\begin{array}{lcl}
f_n(x) &=& f_n(-1-y) \\
    &=& 3(1+y)^n +n(1-y)-3-n(1+y)^2 \\
    &=& 3(1+y)^n -ny^2-3ny-3 \\
    &=& 3(\sum_{j=0}^n \binom{n}{j}y^j)-ny^2-3ny-3 \\
    &=& \frac{n}{2}(3(n-2)+1)y^2+3(\sum_{j=3}^n \binom{n}{j}y^j) \geq 0
\end{array}
$$
If $x\in [-1,2]$, we have $3(1+x^n) \geq 0 \geq n(x^2-x-2)$ so $f_n(x)\geq 0$.
Finally, suppose that $x\geq 2$. We can then write $x=2+z$ with $z\geq 0$, and then
$$
\begin{array}{lcl}
f_n(x) &=& f_n(2+z) \\
    &=& 3(2+z)^n +n(4+z)-3-n(2+z)^2 \\
    &=& 3(2+z)^n -nz^2-3nz-3 \\
    &=& 3(\sum_{j=0}^n \binom{n}{j}2^{n-j}y^j)-nz^2-3nz-3 \\
    &=& 3(2^n-1)+3n(2^{n-1}-1)z+n((n-1)2^{n-3}-1)z^2+ 3(\sum_{j=3}^n \binom{n}{j}2^{n-j}y^j) \geq 0
\end{array}
$$
This finishes the proof.
