Prove $x^4 + x^2 +1$ is always greater than $x^3 + x$

Observe that $P>Q$ for $x=-1$. Now if $x \neq -1$, then $$P-Q=1-x+x^2-x^3+x^4\overbrace{=}^{\text{geometric series}}\frac{x^5+1}{x+1}.$$ If $x>-1$, then both numerator and denominator are positive, thus $P-Q>0$.

If $x<-1$, then both numerator and denominator are negative, still $P-Q>0$.

So $P>Q$ for all $x$.

Consider $$f(x)=(x^4+x^2+1)-(x^3+x).$$ You want to prove that $f(x)>0$ for all $x$. There are several approaches. You could factor $f(x)$. Or you could write $$f(x)=\frac{x^4+(x^2-x)^2+(x-1)^2+1}2.$$

Let

$$H=P-Q=x^4-x^3+x^2-x+1$$

As you have already one the case $x<0$, we will prove the case $x\geq 0$. Clearly, $H(0)=1>0$. For $x\in (0,1]$, we know

$$1\geq x$$

$$x^2\geq x^3$$

This implies

$$H=x^4+(x^2-x^3)+(1-x)>0$$

For $x>1$, we know

$$x^4>x^3$$

$$x^2>x$$

This implies

$$H=(x^4-x^3)+(x^2-x)+1>0$$

and we are done as $H$ has no real roots and $H(0)>0$.