Sonia Ayers

2022-07-03

Determining for what values is a system of inequalities true.

$\begin{array}{}\text{(1)}& 8{A}^{2}\alpha +44A\alpha +60\alpha >3{A}^{2}{\alpha}^{2}+12{A}^{2}+18A{\alpha}^{2}+40A+27{\alpha}^{2}+40\end{array}$

$\begin{array}{}\text{(2)}& A>0\end{array}$

$\begin{array}{}\text{(3)}& \alpha >1\end{array}$

Is there a procedure that determines for what value range(s) of $\alpha $ inequalities (1) and (2) are satisifed?

$\begin{array}{}\text{(1)}& 8{A}^{2}\alpha +44A\alpha +60\alpha >3{A}^{2}{\alpha}^{2}+12{A}^{2}+18A{\alpha}^{2}+40A+27{\alpha}^{2}+40\end{array}$

$\begin{array}{}\text{(2)}& A>0\end{array}$

$\begin{array}{}\text{(3)}& \alpha >1\end{array}$

Is there a procedure that determines for what value range(s) of $\alpha $ inequalities (1) and (2) are satisifed?

alomjabpdl0

Beginner2022-07-04Added 12 answers

LHS is $\alpha (8{A}^{2}+44A+60)$ and RHS is ${\alpha}^{2}(3{A}^{2}+18A+27)+(12{A}^{2}+40A+40)$. And as all components are positive:

So $\alpha (8{A}^{2}+44A+60)>{\alpha}^{2}(3{A}^{2}+18A+27)+(12{A}^{2}+40A+40)\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$

$\frac{8{A}^{2}+44A+60}{3{A}^{2}+18A+27}-\frac{1}{\alpha}\frac{12{A}^{2}+40A+40}{3{A}^{2}+18A+27}>\alpha $

So this will be false whenever $\alpha \ge \frac{8{A}^{2}+44A+60}{3{A}^{2}+18A+27}-\frac{1}{\alpha}\frac{12{A}^{2}+40A+40}{3{A}^{2}+18A+27}$ which will include (but not be restricted to) whenever $\alpha \ge \frac{8{A}^{2}+44A+60}{3{A}^{2}+18A+27}$.

For any $A>0$ we can always find $\alpha \ge \frac{8{A}^{2}+44A+60}{3{A}^{2}+18A+27}$ is which case

$\alpha (8{A}^{2}+44A+60)<{\alpha}^{2}(3{A}^{2}+18A+27)+(12{A}^{2}+40A+40)$

So $\alpha (8{A}^{2}+44A+60)>{\alpha}^{2}(3{A}^{2}+18A+27)+(12{A}^{2}+40A+40)\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$

$\frac{8{A}^{2}+44A+60}{3{A}^{2}+18A+27}-\frac{1}{\alpha}\frac{12{A}^{2}+40A+40}{3{A}^{2}+18A+27}>\alpha $

So this will be false whenever $\alpha \ge \frac{8{A}^{2}+44A+60}{3{A}^{2}+18A+27}-\frac{1}{\alpha}\frac{12{A}^{2}+40A+40}{3{A}^{2}+18A+27}$ which will include (but not be restricted to) whenever $\alpha \ge \frac{8{A}^{2}+44A+60}{3{A}^{2}+18A+27}$.

For any $A>0$ we can always find $\alpha \ge \frac{8{A}^{2}+44A+60}{3{A}^{2}+18A+27}$ is which case

$\alpha (8{A}^{2}+44A+60)<{\alpha}^{2}(3{A}^{2}+18A+27)+(12{A}^{2}+40A+40)$

Shea Stuart

Beginner2022-07-05Added 4 answers

Alternative

$8{A}^{2}\alpha +44A\alpha +60\alpha >3{A}^{2}{\alpha}^{2}+12{A}^{2}+18A{\alpha}^{2}+40A+27{\alpha}^{2}+40\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$

$0>-3{A}^{2}{a}^{2}>(12{A}^{2}+18A{\alpha}^{2}+40A+27{\alpha}^{2}+40)-(44A\alpha +60\alpha )$

It's pretty clear that if we take $A$ and $\alpha $ large enough we can find values where the RHS is positive.

For example: If we let $18A{\alpha}^{2}>44A\alpha $ or in other words let $\alpha >\frac{44}{18}=\frac{22}{9}$ we have:

$(12{A}^{2}+18A{\alpha}^{2}+40A+27{\alpha}^{2}+40)-(44A\alpha +60\alpha )>$

$(12{A}^{2}+44A\alpha +40A+27\cdot \frac{22}{9}\alpha +40)-(44A\alpha +60\alpha )=$

$12{A}^{2}+40A+66\alpha +40)-60\alpha =$

$12{A}^{2}+40A+6\alpha +40$

Which clearly positive.

$8{A}^{2}\alpha +44A\alpha +60\alpha >3{A}^{2}{\alpha}^{2}+12{A}^{2}+18A{\alpha}^{2}+40A+27{\alpha}^{2}+40\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$

$0>-3{A}^{2}{a}^{2}>(12{A}^{2}+18A{\alpha}^{2}+40A+27{\alpha}^{2}+40)-(44A\alpha +60\alpha )$

It's pretty clear that if we take $A$ and $\alpha $ large enough we can find values where the RHS is positive.

For example: If we let $18A{\alpha}^{2}>44A\alpha $ or in other words let $\alpha >\frac{44}{18}=\frac{22}{9}$ we have:

$(12{A}^{2}+18A{\alpha}^{2}+40A+27{\alpha}^{2}+40)-(44A\alpha +60\alpha )>$

$(12{A}^{2}+44A\alpha +40A+27\cdot \frac{22}{9}\alpha +40)-(44A\alpha +60\alpha )=$

$12{A}^{2}+40A+66\alpha +40)-60\alpha =$

$12{A}^{2}+40A+6\alpha +40$

Which clearly positive.

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