How one can show that this polynomial.


is negative for all integer $k>1$ and real $x>2$.

I have no idea to start.


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  • $\begingroup$ @greedoid: No it is $k-1$. $\endgroup$ – DER Oct 10 '18 at 15:59
  • $\begingroup$ Any reason why it should be $x>5$ specifically? desmos.com/calculator/esk3my2jfc $\endgroup$ – Mohammad Zuhair Khan Oct 10 '18 at 16:07
  • $\begingroup$ That will likely become clear in the answer $\endgroup$ – HackerBoss Oct 10 '18 at 16:08
  • $\begingroup$ @HackerBoss I only meant that because it seems that for $k>1,$ $-2+2x^{2k+1}-x^{3k+2}+x^{k-1}$ is negative for all values $x>1.18.$ $\endgroup$ – Mohammad Zuhair Khan Oct 10 '18 at 16:12
  • $\begingroup$ @MohammadZuhairKhan: How you get this result $\endgroup$ – DER Oct 10 '18 at 16:12

For $x>2$ we have $x^{k+1} >4x^{k-1}>x^{k-1}$

so we have:

$$x^{2k+1}(2-x^{k+1})-(2-x^{k-1})<x^{2k+1}(2-x^{k-1})-(2-x^{k-1})= \underbrace{(x^{2k+1}-1)}_{>0}\underbrace{(2-x^{k-1})}_{<0}$$


Well I can change all the signs and assert that the result is positive. If I multiply by $x$ I don't change the sign, because $x$ is positive and I then put $y=x^k$ to simplify the expression. Then I am looking at $$x^3y^3-2x^2y^2-y+2\gt 0$$

I am then guessing that $-2$ is irrelevant once $x$ and $k$ are large, so I drop it and divide through by $y$ (positive) and I expect to find $$x^3y^2-2x^2y-1\gt 0$$ perhaps with some extra cases to consider. Now I drop the $-1$ as small in relation to the other terms and divide through by $x^2y$ (positive) and consider $$xy-2$$

Now with $k\gt 1$ and $x\gt 5$ we have also $y\gt 5$ so that $$xy-2\gt23$$

Now we see how to reverse the process with crude estimates.

Multiply through by $x^2y^\gt 1$ (very crude estimate) and subtract $1$ to obtain $$x^3y^2-2x^2y-1\gt 23x^2y-1\gt 22$$

Multiply by $y\gt 1$and add $2$ to obtain $$x^3y^3-2x^2y^2-y+2\gt22y+2\gt 24\gt 0$$

which is equivalent to the original inequality. As you see, only very crude estimates are required. I used $y$ to see if there were any obvious squares hiding in he background.

  • 1
    $\begingroup$ This is a beautiful answer. Thanks. $\endgroup$ – Surb Oct 10 '18 at 20:01

Note that $$-2+2x^{2k+1}-x^{3k+2}+x^{k-1}=-2+\left(2x^{k+2}-x^{2k+3}+1\right)x^{k-1}$$ and $$2x^{k+2}-x^{2k+3}+1=1+\left(2-x^{k+1}\right)x^{k+2}$$ If $x > 5$, and $k>1$, then $2-x^{k+1}<-1$, $x^{k+2}>1$, and $x^{k-1}>1$, so that $$-2+\left(1+\left(2-x^{k+1}\right)x^{k+2}\right)x^{k-1}<0$$ Informally, $$-2+(1+(<-1)(>1))(>1)\rightarrow-2+(1+(<-1))(>1)\rightarrow-2+(<0)(>1)\rightarrow-2+(<0)<0$$


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