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Prove that $$x^8-x^5-x^4+x^2+x>-\frac{1}{3}$$ for all real $x$

My partial proof

First I tried to simplify and factorise $3x^8-3x^5-3x^4+3x^2+3x+1$ but factoring or simplification didn't work. The polynomial is irreducible over $\Bbb Q$. I tried to restrict $x$ in some intervals. For example $0<x<1$ or $x>1$ or $x<0$. I obtained $$x^5(x^3-1)-x(x^3-1)+x^2+\frac{1}{3}>0$$ or $$(x^5-x)(x^3-1)+x^2+\frac{1}{3}>0$$ or $$x(x^4-1)(x^3-1)+x^2+\frac{1}{3}>0$$ or $$x(x^2-1)(x^2+1)(x-1)(x^2+x+1)+x^2+\frac{1}{3}>0$$ or $$x(x-1)(x+1)(x^2+1)(x-1)(x^2+x+1)+x^2+\frac{1}{3}>0$$ or $$x(x-1)^2(x+1)(x^2+1)(x^2+x+1)+x^2+\frac{1}{3}>0$$

So it remains to prove the inequality for $x(x+1)<0$ which gives $-1<x<0$.

hardmath
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  • Also not directly successful: Multiply with 2 and use $x^8$ to get complete squares absorbing the 5th and 4th powers $$ x^2(x^3-1)^2+(x^4-1)^2+(x+1)^2\overset{?}>2-\frac23 $$ – Lutz Lehmann Oct 11 '23 at 18:58

2 Answers2

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Just playing around led to $$x^8 - x^5 - x^4 + x^2 + x = \left (x^4 - \frac x2 - \frac 12 \right )^2 + \frac 34 \left(x + \frac 13 \right)^2 - \frac 13.$$

WimC
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    Thanks but I like answers not use computer softwares – hardmath Oct 11 '23 at 19:03
  • @hardmath then this is the answer you should like. $x^5$ being the highest order term after $x^8$ immediately suggests completing the square with $(x^4-\frac x2-...)^2.$ – dezdichado Oct 11 '23 at 19:27
  • @hardmath see if you can write $x^4 + x^3 - 2 x^2 - 3x - 1$ as a difference of squares. The question posted a few hours ago concerned $x^4 + x^3 - 2 x^2 - 3ax - a^2$ where they had restrictions on $a$ – Will Jagy Oct 11 '23 at 20:01
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Another way .

In this answer, we apply a method that only works with Discriminants. ( also see here and here )

The main idea of ​​the method is as follows :

Let $\small{\begin{align}P(x)=x^8-x^5-x^4+x^2+x+\frac 13\end{align}}$ . We transform the original polynomial into the quadratic-like polynomial with respect to $\color{#c00}{x}$ :

$$ \small{\begin{align}\overbrace{\underbrace{(x^6-x^3+1)}_{>\thinspace 0,\thinspace \forall x\thinspace\in\thinspace \mathbb R}}^{\Delta_{x^3}\thinspace <\thinspace 0}\color{#c00}{x^2}-(x^3-1)\color{#c00}{x}+\frac 13\end{align}} $$

We observe that, $P(-1)>0$ . Then, by substituting $x^3=u$, we determine the discriminant $\Delta_{\color{#c00}{x}}$ :

$$ \begin{align}\Delta_{\color{#c00}{x}}&=(u-1)^2-\frac 43(u^2-u+1)\\ &=-\frac 13(u+1)^2<0,\thinspace \forall u\in\mathbb R_{≠-1}\end{align} $$

Finally, since $\Delta_{\color{#c00}{x}}<0,\thinspace \forall x\in\mathbb R$ this means that $P(x)>0,\thinspace \forall x\in\mathbb R$ .

lone student
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    It is interesting. Alternatively (without discriminant), we may use AM-GM; it suffices to prove that $4 \cdot (x^6 - x^3 + 1) \cdot \frac13 \ge (x^3 - 1)^2$. (In my opinion, it is more natural than discriminant.) By the way, the point is that, in general, is it easy to write the expression in the form of $A(x)x^2 - B(x)x + C(x) \ge 0$ satisfying $4\cdot A(x) \cdot C(x) - [B(x)]^2 \ge 0$. – River Li Oct 20 '23 at 04:17
  • For example, can you prove this one: $16x^9-56x^6+64x^5+65x^3-64x^2+7 > 0$ for all $x\in [0, 1]$? It is one step in my old answer. – River Li Oct 20 '23 at 05:21
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    I think, I'm not strong enough to write the proof in your example . – lone student Oct 27 '23 at 18:39