khong mat tinh tong quat ta gia su $a\geq b\geq c$$\frac{1}{(a-b)^{2}}+\frac{1}{(b-c)^{2}}\geq \frac{2}{(a-b)(b-c)}\geq \frac{8}{(a-c)^{2}}$
$\leftrightarrow A= \frac{1}{(a-b)^{2}}+\frac{1}{(b-c)^{2}}+\frac{1}{(c-a)^{2}}\geq \frac{9}{(c-a)^{2}}$
ta co: $(c-a)^{2}\leq 2(a^{2}+c^{2})\leq 2(a^{2}+b^{2}+c^{2})\leftrightarrow A\geq \frac{9}{2(a^{2}+b^{2}+c^{2})}$
$\rightarrow P\geq \frac{9}{2}$
dau bang xay ra khi \begin{cases}a=-c \\ b=0 \end{cases}
( hoac ta co the hoan doi cac vi tri a,b,c, cho nhau)