How one can show that this polynomial is negative

Question

How one can show that this polynomial.

$$-2+2x^{2k+1}-x^{3k+2}+x^{k-1}$$

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

I have no idea to start.

Answer 1

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

so we have:

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

Answer 2

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.

Answer 3

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, $$-2+(1+(<-1)(>1))(>1)\rightarrow-2+(1+(<-1))(>1)\rightarrow-2+(<0)(>1)\rightarrow-2+(<0)<0$$