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I am trying to prove an inequality using the $pqr$ method. When factoring a high degree polynomial I face the following inequality: $2a^4+4a^3 - 4a^2-a+2>0$ when $0 < a < 1$.

I tried to split it into two inequalities each is $\ge 0$, but I am still wrestling with it now without the light at the end of the tunnel. Any hint is helpful. I graphed it using wolfram alpha and the graph stays above the $x$ axis. Thus the inequality is valid. WY.

MathFail
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Wang YeFei
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  • Can you show/clarify what you did? EG You stated a single-variable inequality, but you referenced the pqr method. – Calvin Lin May 05 '23 at 03:09
  • The inequality above is a factor of another $6$-degree polynomial that is the result of the $pqr$ method used to prove symmetric inequality in $3$ variables when there is a constraint such as $ab+bc+ ca = 6$ for example. – Wang YeFei May 05 '23 at 04:00
  • I think it would be better to prove that the following polynomial $>0 \forall a \geq 0$ – Hersh May 05 '23 at 04:09
  • Yes I can see it true quite clearly now. But often times when a constrain is imposed on it, it makes it more interesting. – Wang YeFei May 05 '23 at 04:13
  • I don't know what is the "pqr method". Could you give a reference ? – Jean Marie May 05 '23 at 05:39

4 Answers4

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$$\begin{align}\color{red}{2a^4}+\color{blue}{4a^3}-4a^2-a+2&\ge \color{red}{0a^4}+\color{blue}{3a^3}-4a^2-a+2=(a-1)^2(3a+2)>0\end{align}$$

MathFail
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I just came up with my own answer: For $a > \dfrac{1}{2}$ we have: $f(a) = a(4a^2 - 1) + 2(a^4 - 2a^2+1) = a(4a^2 -1) + 2(a^2 - 1)^2 > 0$, and if $0 < a < \dfrac{1}{2}$ then $f(a) > 2a^4+4a^3 - 1 - a + 2 = 2a^4+4a^3 -a + 1 > 1 - a > 0$ since $0 < a < \dfrac{1}{2}$. Thus we proved the inequality.The case $a = \dfrac{1}{2}$ is clear.

Wang YeFei
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Because $$2a^4+4a^3-4a^2-a+2=2(a^2+a-1)^2+a(3-2a)>0.$$

Michael Rozenberg
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I do not know what is $pqr$, but I would want to suggest you the following way :

If $\thinspace 0<a<1$, then :

$$ \begin{align}P(a)&>\color{red}{0}+\color{red}{4a^4}-4a^2\color{red}{-1}+2\\ &=\left(2a^2-1\right)^2\\ &≥0\thinspace .\end{align} $$

lone student
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