Prove that $x^{16}-x^{11}+x^6-x+1>0$ for $x\in R$.

So I thought of something like this: $$x^{10}(x^6-x)+x^6-x>-1$$ $$(x^{10}+1)(x^6-x)>-1$$

But it seems to not be too much of help. While the first bracket is always positive, the latter can't be (or I don't know how to do this) easily transformed to a form showing that it's positive. I know it can be solved by thinking about three cases (for x>1, x<0 and $x\in <0,1>$), but it involves transforming the initial inequality three times each time basing rather on our skill with such problems (I mean - someone who hasn't worked on such previously would have a hard time getting the inequality to a form suitable for showing every case) rather than observations on-the-spot but can this be solved in some easier manner?

  • $\begingroup$ Hint: when $x=-1$, then $1+x(1+x^{10})(x^5-1)=1-2\cdot(-2)=5 > 0$. So what can you say about $x^6-x=x(x^5-1)=x(x-1)(x^4+x^3+x^2+x+1)$ (especially if you knew tht the last, quartic, factor had no real roots)? You should be able to conclude that the only interval to worry about is $x\in(0,1)$, which is pretty easy to dispense with. As to the quartic, its roots belong to the roots of $x^5-1$ which form a regular pentagon on the unit circle in the complex plane. $\endgroup$ – bgins Apr 29 '12 at 16:15
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    $\begingroup$ For $|x| < 1$ you could also use $-1 < \frac{-x}{1 + x^5} = -x + x^6 - x^{11} + x^{16} - \ldots < x^{16} - x^{11} + x^6 - x$. $\endgroup$ – TMM Apr 29 '12 at 16:21
  • $\begingroup$ That doesn't make it easier but still another way out so thank you :) $\endgroup$ – Straightfw Apr 29 '12 at 17:31

Factoring as you did above, we are looking at $$(x^{10}+1)(x^6-x)+1.$$ The term $x^{10}+1$ is always $\geq 0$, and $x^6-x\geq 0$ when $x\leq 0$ or when $x\geq 1$. This means we have verified the inequality holds everywhere except for $x\in(0,1)$. To deal with this interval, group terms and write the polynomial as $$ (1-x)+(x^6-x^{11})+(x^{16}).$$ For $x\in (0,1)$ each of these terms is positive, which allows us to conclude that the inequality holds for all $x\in\mathbb{R}$.

  • $\begingroup$ OK, so we need this one additional grouping to show this. Thank you :) $\endgroup$ – Straightfw Apr 29 '12 at 17:30

There is no problem when $x$ is negative.

Now deal with $x\ge 0$. It seems natural to multiply by $x^5+1$. We get $x^{21}+x^5-x+1$. If $0\le x\lt 1$, already $-x+1$ is positive. If $x \ge 1$, then $x^5-x \ge 0$, so $x^{21}+x^5-x+1$ is positive. Multiplying works in the same way with longer polynomials of similar structure.


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