Find all triples of positive real numbers $(a, b, c)$ so that the expression

$$
M=\frac{(a+b)(b+c)(a+b+c)}{a b c}
$$

gets its least value.
The expression $M$ is homogeneous, therefore we can assume that $a b c=1$. We set $s=a+c$ and $p=a c$ and using $b=\frac{1}{a c}$, we get

$$
M=\left(a+\frac{1}{a c}\right)\left(\frac{1}{a c}+c\right)\left(a+\frac{1}{a c}+c\right)=\left(a+p^{-1}\right)\left(c+p^{-1}\right)\left(s+p^{-1}\right)
$$

Expanding the right-hand side we get

$$
M=p s+\frac{s^{2}}{p}+1+\frac{2 s}{p^{2}}+\frac{1}{p^{3}}
$$

Now by $s \geq 2 \sqrt{p}$ and setting $x=p \sqrt{p}>0$ we get

$$
M \geq 2 x+5+\frac{4}{x}+\frac{1}{x^{2}}
$$

We will now prove that

$$
2 x+5+\frac{4}{x}+\frac{1}{x^{2}} \geq \frac{11+5 \sqrt{5}}{2} \text {. }
$$

Indeed, the latter is equivalent to $4 x^{3}-(5 \sqrt{5}+1) x^{2}+8 x+2 \geq 0$, which can be rewritten as

$$
\left(x-\frac{1+\sqrt{5}}{2}\right)^{2}(4 x+3-\sqrt{5}) \geq 0
$$

which is true.

Remark: Notice that the equality holds for $a=c=\sqrt{p}=\sqrt[3]{\frac{1+\sqrt{5}}{2}}$ and $b=\frac{1}{a c}$.