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 $x\lt1$ we have $x(x^2+1)\lt(x^2+1)$, since $x^2+1\gt0$, and therefore
$$x^3+x=x(x^2+1)\lt x^2+1\le x^4+x^2+1$$
For $x\ge1$, we have $x^3+x\le x(x^3+x)$, and therefore
$$x^3+x\le x(x^3+x)=x^4+x^2\lt x^4+x^2+1$$
Thus $x^3+x\lt x^4+x^2+1$ for all $x$.
