For $a,b,c$ are real numbers such that $ab+bc+ca>0$ with $a+b+c=1$,find min
$$P=\frac{2}{|a-b|}+\frac{2}{|b-c|}+\frac{2}{|c-a|}+\frac{5}{\sqrt{ab+bc+ca}}$$

（choisiwon，Japan）

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=539142
Solution (thanks tian275461's hint)
We will prove that
$$P\geq 10\sqrt{6}$$
Without loss of generally,we can assume that $a>b>c$
by AM-GM inequality
$$\frac{2}{a-b}+\frac{2}{b-c}\geq \frac{8}{a-c}$$
which leads
$$P\geq \frac{10}{a-c}+\frac{5}{\sqrt{ab+bc+ca}}\geq \frac{4}{\frac{a-c}{10}+\frac{\sqrt{ab+bc+ca}}{5}}$$
Thus,it's suffice to prove
$$\frac{a-c}{10}+\frac{\sqrt{ab+bc+ca}}{5}\leq \frac{4}{10\sqrt{6}}=\frac{1}{15}\sqrt{6}$$
Or
$$(a-c)+2\sqrt{(a+c)-(a+c)^2+ac}\leq \frac{2}{3}\sqrt{6}$$
Now,Using Cauchy-Schwarz and AM-GM inequality,we have
$$\begin{align*}  (a-c)+2\sqrt{(a+c)-(a+c)^2+ac}&\leq \sqrt{2[(a-c)^2+4(a+c)-4(a+c)^2+4ac]}\\ &=\sqrt{2(a+c)[4-3(a+c)]}\\ &=\frac{\sqrt{2}}{\sqrt{3}}\sqrt{3(a+c)[4-3(a+c)]}\\ &\leq \frac{2}{3}\sqrt{6} \end{align*}$$
Done!