# 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.

• Consider the min. value of both sides. It may be a simple approach. – NoChance Sep 28 '19 at 12:34

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.

• This is a good proof. Minor note: one can simplify things slightly if one notes that when $n$ is even $f'_n(x)$ is positive when $x \geq 0$ so one only needs to look at $[-1,1]$ in the second case and don't need to look at the last interval at all. – JoshuaZ Sep 28 '19 at 12:36