WLOG b=1 a+c=x ac=y then you get that the expression minus some constant is (1+x^2)/y + x and from here is is obvious that x should be minimal so a=c and then use derivative to do the rest

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$\definecolor{A}{RGB}{0,90,255}\color{A}\fbox{Solution}$ Cauchy- Schwarz and AM-GM $$M\ge\frac{(b+\sqrt{ac})^2(b+2\sqrt{ac})}{abc}=\frac{(x+1)^2(x+2)}{x}$$where $x=\frac{b}{\sqrt{ac}}>0$ with equality iff $a=c$. $$x>0\implies (x+1)^2(x+2)-x\frac{(\sqrt{5}+1)^3(\sqrt{5}+3)}{16}=(x+3+\sqrt{5})\left(x+\frac{1-\sqrt{5}}{2}\right)^2\ge0.$$Therefore $$M\ge \frac{(\sqrt{5}+1)^3(\sqrt{5}+3)}{16}$$with equality iff$$\definecolor{A}{RGB}{100,0,255}\color{A}b=a\cdot \frac{\sqrt{5}-1}{2}= c\cdot \frac{\sqrt{5}-1}{2}.\blacksquare$$#1848

Since the expression is homogeneous, assume $c=1$. Then $M = \frac{b^2}{a} + \frac{2b}{a} + \frac{a}{b} + a + \frac{1}{a} + 2b + \frac{1}{b} + 3$. Let $t = \frac{a}{b}$. Then $M = \frac{b}{t} + \frac{2}{t} + t + bt + \frac{1}{bt} + 2b + \frac{1}{b} + 3$. Let $f(t) = \frac{b}{t} + \frac{2}{t} + t + bt + \frac{1}{bt} + 2b + \frac{1}{b} + 3$. Then, $f'(t) = b + 1 - \frac{2}{t^2} - \frac{b}{t^2} - \frac{1}{bt^2} = 0 \iff t = \sqrt{\frac{b+1}{b}}$. So minima occurs when $t = \sqrt{\frac{b+1}{b}}$. Then, $M = 2\sqrt{\frac{(b+1)^3}{b}} + 2b + \frac{1}{b} + 3$. Now let $g(b) = 2\sqrt{\frac{(b+1)^3}{b}} + 2b + \frac{1}{b} + 3$. We have $g'(b) = 0 \iff b = \frac{\sqrt{5} - 1}{2}$. So, $g(b) \geq g(\frac{\sqrt{5} - 1}{2}) = \frac{11 + 5\sqrt{5}}{2} \implies M \geq \frac{11 + 5\sqrt{5}}{2}$. Equality is achieved when $a : b : c = 1 : \frac{\sqrt{5} - 1}{2} : 1$. $\square$ @below Yeah edited.

hakN wrote: Equality is achieved when $a : b : c = \frac{\sqrt{5} - 1}{2} \cdot \sqrt{\frac{\sqrt{5} + 3}{2}} : \frac{\sqrt{5} - 1}{2} : 1$. $\square$ $$\frac{\sqrt{5} - 1}{2} \cdot \sqrt{\frac{\sqrt{5} + 3}{2}}=1$$

Mikeglicker wrote: WLOG b=1 a+c=x ac=y then you get that the expression minus some constant is (1+x^2)/y + x and from here is is obvious that x should be minimal so a=c and then use derivative to do the rest Could you explain it please?

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