# Determining for what values is a system of inequalities true. <mtable displaystyle="true">

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?
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alomjabpdl0
LHS is $\alpha \left(8{A}^{2}+44A+60\right)$ and RHS is ${\alpha }^{2}\left(3{A}^{2}+18A+27\right)+\left(12{A}^{2}+40A+40\right)$. And as all components are positive:
So $\alpha \left(8{A}^{2}+44A+60\right)>{\alpha }^{2}\left(3{A}^{2}+18A+27\right)+\left(12{A}^{2}+40A+40\right)\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$
$\frac{8{A}^{2}+44A+60}{3{A}^{2}+18A+27}-\frac{1}{\alpha }\frac{12{A}^{2}+40A+40}{3{A}^{2}+18A+27}>\alpha$
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
$\alpha \left(8{A}^{2}+44A+60\right)<{\alpha }^{2}\left(3{A}^{2}+18A+27\right)+\left(12{A}^{2}+40A+40\right)$
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Shea Stuart
Alternative
$8{A}^{2}\alpha +44A\alpha +60\alpha >3{A}^{2}{\alpha }^{2}+12{A}^{2}+18A{\alpha }^{2}+40A+27{\alpha }^{2}+40\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}$
$0>-3{A}^{2}{a}^{2}>\left(12{A}^{2}+18A{\alpha }^{2}+40A+27{\alpha }^{2}+40\right)-\left(44A\alpha +60\alpha \right)$
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:
$\left(12{A}^{2}+18A{\alpha }^{2}+40A+27{\alpha }^{2}+40\right)-\left(44A\alpha +60\alpha \right)>$
$\left(12{A}^{2}+44A\alpha +40A+27\cdot \frac{22}{9}\alpha +40\right)-\left(44A\alpha +60\alpha \right)=$
$12{A}^{2}+40A+66\alpha +40\right)-60\alpha =$
$12{A}^{2}+40A+6\alpha +40$
Which clearly positive.