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.
 A: 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 $x$.
A: 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
$$f(x)=\frac{x^4+(x^2-x)^2+(x-1)^2+1}2.$$
A: 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 $H(0)>0$.
A: 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 $x$.
A: Both functions are increasing on $[0,1]$.
$$
\begin{alignat}{4}
\text{On $[0,1/2]$,}\quad && Q(x)&<Q(1/2)=\frac58&<1&=P(0)&<P(x).\\
\text{On $[1/2,4/5]$,}\quad && Q(x)&<Q(4/5)=\frac{164}{125}&<\frac{21}{16}&=P(1/2)&<P(x).\\
\text{On $[4/5,1]$,}\quad && Q(x)&<Q(1)=2&<\frac{1281}{625}&=P(4/5)&<P(x).
\end{alignat}
$$
A: 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.
A: For $x\leq0$ it's obvious.
But for $x>0$ we obtain:
$$x^4+x^2+1-(x^3+x)=x^4-2x^3+3x^2-2x+1+x^3+2x^2+x=$$
$$=(x^2-x+1)^2+x(x+1)^2>0.$$
A: 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.
