In resticted domain , Applying the Cauchy-Schwarz's inequality

Question

I see this problem a few days ago.

$$ a, b, c \in [\alpha , \beta] $$ prove that $$ 9 \le (a+b+c)\biggl( \frac {1}{a} + \frac{1} {b} + \frac{1} {c} \biggr)\le \frac{(2 \alpha + \beta )(\alpha + 2\beta )}{ \alpha \beta}$$

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 $

Answer 1

I hope you mean that $\alpha>0$.

Let $f(a,b,c)=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$.

Thus, $f$ is a convex function of $a$, of $b$ and of $c$ (for example, $\frac{\partial f}{\partial a^2}=\frac{2(b+c)}{a^3}>0$), which says that $$\max_{\{a,b,c\}\subset[\alpha,\beta]}f=\max_{\{a,b,c\}\subset\{\alpha,\beta\}}f=f(\alpha,\alpha,\beta)=\frac{(2\alpha+\beta)(\alpha+2\beta)}{\alpha\beta}$$ because $f(\alpha,\alpha,\alpha)=f(\beta,\beta,\beta)=9$ and by AM-GM $$\frac{(2\alpha+\beta)(\alpha+2\beta)}{\alpha\beta}=2\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)+5\geq4+5=9.$$