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.

Proving that $x^4 + x^2 +1$ is always above $x^3 + x$ si the same as proving that
$$f(x)=x^4-x^3+x^2-x+1$$ is always positive. We have
$$f'(x)=4 x^3-3 x^2+2 x-1$$ has only one real root; using the hyprbolic method, we find that its real root is given by
$$x_*=\frac{1}{12} \left(3+2 \sqrt{15} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(3
   \sqrt{\frac{3}{5}}\right)\right)\right)\approx 0.605830$$ and the second derivative test reveals that $x_*$ corresponds to a minimum.
$$f(x_*)\approx 0.673553$$ So, it is always true.
