5
$\begingroup$

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?

  • 0
    That doesn't make it easier but still another way out so thank you :)2012-04-29

2 Answers 2

15

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 $x\in\mathbb{R}$.

  • 0
    OK, so we need this one additional grouping to show this. Thank you :)2012-04-29
1

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.