(non treated in course) Determine the minimum of the expression

$$
\frac{2}{|a-b|}+\frac{2}{|b-c|}+\frac{2}{|c-a|}+\frac{5}{\sqrt{a b+b c+c a}}
$$

under the conditions that $a b+b c+c a>0, a+b+c=1$ and $a, b, c$ are distinct.
For symmetry reasons, we reduce to the case $a > b > c$. Then the expression is

$$
\begin{aligned}
A & =\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}+\frac{5}{\sqrt{a b+b c+c a}} \\
& \geqslant \frac{10}{c-a}+\frac{5}{\sqrt{a b+b c+c a}} \text{ since } \frac{1}{m}+\frac{1}{n} \geqslant \frac{4}{m+n} \\
& =10\left(\frac{1}{c-a}+\frac{1}{2 \sqrt{a b+b c+c a}}\right) \\
& \geqslant \frac{20 \sqrt{2}}{\sqrt{(c-a)^{2}+4(a b+b c+c a)}} \text{ since } \frac{1}{m}+\frac{1}{n} \geqslant \frac{2 \sqrt{2}}{\sqrt{m^{2}+n^{2}}} \\
& =\frac{20 \sqrt{2}}{\sqrt{(a+c)^{2}+4 b(a+c)}}=\frac{20 \sqrt{2}}{\sqrt{(1-b)(1+3 b)}}
\end{aligned}
$$

We bound this expression from below by $10 \sqrt{6}$, the bound being achieved for $b=1 / 3$ and $a+c=2 / 3$ satisfying $(c-a)^{2}=4(a b+b c+c a)$, i.e., $a=\frac{2+\sqrt{6}}{6}$ and $c=\frac{2-\sqrt{6}}{6}$.