Áp dụng BĐT : $x^2+y^2+z^2\ge \frac{1}{3}(x+y+z)^2$Ta có
$P=(\frac{1}{a}+\frac{2}{b+c}+\frac{3}{a+b+c})^2+ .....$
$\ge \frac{1}{3}[(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+2(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})+\frac{9}{a+b+c}]^2$
Mà
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{9}{a+b+c}$
$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge \frac{9}{2a+2b+2c}$
$\Rightarrow P\ge \frac{1}{3}(\frac{27}{a+b+c})^2=81.\frac{3}{(a+b+c)^2}\ge \frac{81}{a^2+b^2+c^2}$