Minimum of $\left(\frac{1}{a}+\frac{2}{b}+\frac{3}{c}\right)$ where $a,b,c$ are the roots of a polynomial

Question

If $a,b,c$ are positive roots of $$x^3-lx^2+mx-48=0$$ Find the minimum value of: $\left(\cfrac{1}{a}+\cfrac{2}{b}+\cfrac{3}{c}\right)$

So:

$$abc=48$$

$$a+b+c=l$$

$$ab+bc+ca=m$$

Through AM$\ge$GM, I found that $$l\ge 6\sqrt[3]{6}$$

I don't think that's useful.

Also, I tried AM$\ge$HM: $$\cfrac{ \cfrac{1}{a}+\cfrac{1}{b}+\cfrac{1}{b}+\cfrac{1}{c}+\cfrac{1}{c}+\cfrac{1}{c} }{6} \ge \cfrac{6} {a+b+b+c+c+c}$$

But it's messing up, because I'm getting an extra $b+2c$ term in the denominator.

Also I don't think Cauchy Shwarz/Lagranges would work here. Any hints would be helpful thanks.

And I'm also unable to find the use of $m$ here.

Answer 1

you are complicating it..

by AM-GM $$\frac{1}{a}+\frac{2}{b}+\frac{3}{c}\ge 3\sqrt[3]{\frac{6}{abc}}=3/2$$ Done!