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I could not find any decent approach to solve this inequality. I would appreciate any help, and input if this is even possible to solve(without a computer).

$$x^8-x^5+x^2-x+1>0$$

Bad at Mathematics
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3 Answers3

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Using the AM-GM inequality, we have $$x^8 + \frac{x^2}{2} \geqslant 2\sqrt{x^8 \cdot \frac{x^2}{2}} = \sqrt{2} \cdot |x^5| \geqslant x^5,$$ and $$1+\frac{x^2}{2} \geqslant 2\sqrt{\frac{x^2}{2}} = \sqrt 2 \cdot |x| \geqslant x$$ Therefore $$x^8+x^2+1> x^5 + x.$$ Done.

Note. The SOS form $$x^{8}+x^{2}+1-x^{5}-x=\left(x^{4}-\frac{x}{2}\right)^{2}+\left(\frac{x}{2}-1\right)^{2}+\frac{x^{2}}{2}>0.$$

nguyenhuyenag
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Hint:

i) For $x \le 0$, you can prove that $f(x) \gt 0$, in fact $\ge 1$.
ii) For $x \ge 1$, again you can prove it holds true.
iii) For $0 \lt x \lt 1$, write $x$ as $\frac{1}{y}$ where $y \gt 1$. You can again prove as (ii).

Math Lover
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Let $f(x)=1\cdot (x^4)^2-(x^4)x+x^2-x+1$

Its discriminant $\displaystyle D =x^2-4(x^2-x+1)$

$\displaystyle D=-3x^2+4x-4=-3\bigg[\bigg(x-\frac{2}{3}\bigg)^2+\frac{32}{9}\bigg]<0$

So we have $f(x)>0$ for all real $x$

Alternate :

$\displaystyle f(x)=x^8-x^5+x^2-x+1$

$\displaystyle f(x)=\frac{1}{2}\bigg[2x^8-2x^5+2x^2-2x+2\bigg]$

$\displaystyle f(x)=\frac{1}{2}\bigg[x^8+\bigg(x^4-x^2\bigg)^2+(x-1)^2+1\bigg]>0$ for all real $x$

jacky
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    Nice. Equivalently, we may use AM-GM: It suffices to prove that $2\sqrt{x^8 \cdot (x^2-x+1)} - x^5 \ge 0 $. Similar idea is used by https://math.stackexchange.com/questions/4785097/prove-that-x8-x5-x4x2x-frac13-for-all-real-x/4785138#comment10184759_4785138 – River Li Nov 01 '23 at 22:52