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.

In other words you want to show that $$x^4-x^3+x^2-x+1>0$$ identically.
This is clearly true if $x=0.$
So for $x\ne 0$ factor out $x^2>0$ to get $$x^2\left(x^2-x+1-1/x+1/x^2\right),$$ and then we only need consider the expression in parentheses. This may be written as $$x^2+1/x^2-(x+1/x)+1,$$ and now we set $u=x+1/x,$ to obtain $$u^2-u-1,$$ whose roots are $$\frac12(1\pm \sqrt 5),$$ so that we only need focus on the interval between these roots. That is this quantity is not positive for $$\frac{1-\sqrt 5}{2}< x+\frac 1x<\frac{1+\sqrt 5}{2}.$$
We now show this to have no solution by considering when the roots of $x+1/x=a$ are not real for some given real value of $a.$ That is, we consider the equation $$x^2-ax+1=0,$$ whose discriminant $a^2-4<0\implies |a|<2.$
We only need show that the roots of $u^2-u-1=0$ are less than $2$ in magnitude, and the proof is complete. But this is easily shown.
