áp dụng Bđt Cauchy Schwarz ta có$2\left ( \sqrt{\frac{2}{(a-b)^2}+\frac{2}{(b-c)^2}}+\frac{1}{|c-a|} \right )+\frac{5}{\sqrt{ab+bc+ca} }\geq \frac{2}{|a-b|}+\frac{2}{|b-c|}+\frac{2}{|c-a|}+\frac{5}{\sqrt{ab+bc+ca} }$
giải sử $a>b>c$ khi đó
$P\geq \frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{a-c}+\frac{5}{\sqrt{ab+bc+ca} }$
ta có $\frac{1}{x}+\frac{1}{y}\geq \frac{4}{x+y}\geq \frac{2\sqrt{2} }{\sqrt{x^2+y^2} }$ với $x,y>0$
$\Rightarrow P\geq \frac{8}{a-c}+\frac{2}{a-c}+\frac{5}{\sqrt{ab+bc+ca} }=\frac{10}{a-c}+\frac{10}{2\sqrt{ab+bc+ca} }\geq \frac{20\sqrt{2} }{\sqrt{(a-c)^2+4(ab+bc+ca} }=\frac{20\sqrt{2} }{\sqrt{(a+c)^2+4b(a+c)} }$
$P\geq \frac{20\sqrt{2} }{\sqrt{(a+c)(a+c+4b)} }=\frac{20\sqrt{2} }{\sqrt{(1-b)(1+3b)} }=\frac{20\sqrt{6} }{\sqrt{(3-3b)(1+3b)} }$
Mặt khác $\sqrt{(3-3b)(1+3b)}\leq \frac{3-3b+1+3b}{2}=2\Rightarrow P\geq 10\sqrt{6} $
Dấu $'='\Leftrightarrow \begin{cases}x3-3b=1+3b \\ a-c=b-c \end{cases} a+b+c=1 \Rightarrow a=\frac{2+\sqrt{6} }{6},b=\frac{1}{3},c=\frac{2-\sqrt{6} }{6}$