Prove $x^4 + x^2 +1$ is always greater than $x^3 + x$ Let's say P is equal to $x^4 + x^2 +1$ and $Q$ is equal to $x^3 + x$.
For $x <0$, $P$ is positive and $Q$ is negative. Hence, in this region, $P>Q$.
For $x=0$, $P>Q$.
Also, for $x = 1$, $P>Q$.
For $x > 1$, I factored out $P$ as $x^2(x^2+1) + 1$ and $Q$ as $x(x^2+1)$. For $x > 1$, $x^2(x^2+1) > x(x^2+1)$, hence $P>Q$.
The part where I have the problem is I can't prove this for the range $0 < x < 1$ without the help of a graphing calculator. Can anyone help?
What I've done in this region so far is:

*

*Prove that $P$ and $Q$ is always increasing in this region,

*The range for $P$ starts from $1 < P < 3$, and

*The range for $Q$ starts from $0 < Q < 2$.

The only thing I need to prove now is that $P$ and $Q$ will not intersect at $0 < x < 1$, but I can't prove this part.

For $x\leq0$ it's obvious.
But for $x>0$ we obtain:
$$x^4+x^2+1-(x^3+x)=x^4-2x^3+3x^2-2x+1+x^3+2x^2+x=$$
$$=(x^2-x+1)^2+x(x+1)^2>0.$$
