# In resticted domain , Applying the Cauchy-Schwarz's inequality I see this problem a few days ago. a, b, c in [alpha , beta] prove that 9 <= (a+b+c)(1/a +1/b +1/c)<= ((2 alpha + beta)(alpha + 2 beta ))/( alpha beta)

In resticted domain , Applying the Cauchy-Schwarz's inequality
I see this problem a few days ago.
$a,b,c\in \left[\alpha ,\beta \right]$
prove that
$9\le \left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le \frac{\left(2\alpha +\beta \right)\left(\alpha +2\beta \right)}{\alpha \beta }$
Left inequality is easy by Cauchy-Schwart's inequality. But Right inequality is difficult to me. How can I approach the Right inequlity?
EDIT : Sorry, I forgot the condition : $\alpha ,\beta >0$
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Dana Simmons
I hope you mean that $\alpha >0$
Let $f\left(a,b,c\right)=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$
Thus, $f$ is a convex function of $a$, of $b$ and of $c$ (for example, $\frac{\mathrm{\partial }f}{\mathrm{\partial }{a}^{2}}=\frac{2\left(b+c\right)}{{a}^{3}}>0$), which says that
$\underset{\left\{a,b,c\right\}\subset \left[\alpha ,\beta \right]}{max}f=\underset{\left\{a,b,c\right\}\subset \left\{\alpha ,\beta \right\}}{max}f=f\left(\alpha ,\alpha ,\beta \right)=\frac{\left(2\alpha +\beta \right)\left(\alpha +2\beta \right)}{\alpha \beta }$
because $f\left(\alpha ,\alpha ,\alpha \right)=f\left(\beta ,\beta ,\beta \right)=9$ and by AM-GM
$\frac{\left(2\alpha +\beta \right)\left(\alpha +2\beta \right)}{\alpha \beta }=2\left(\frac{\alpha }{\beta }+\frac{\beta }{\alpha }\right)+5\ge 4+5=9.$