ta có $3(a^{2}+ b^{2}+ c^{2})=(a+b+c)(a^{2}+ b^{2}+ c^{2})=a^{3}+ b^{3} +c^{3} +a^{2}b+ b^{2}c+ c^{2}a+ ab^{2}+bc^{2}+ca^{2}$ $a^{3}+ ab^{2}\geq2a^{2}b ,b^{3}+bc^{2}\geq 2b^{2}c, c^{3} +ca^{2}\geq 2c^{2}a$
$\Rightarrow 3(a^{2} +b^{2}+c^{2})\geq 3(a^{2}b+ b^{2}c+ c^{2}a)\Rightarrow a^{2}+b^{2}+c^{2} \geq a^{2}b+ b^{2}c+c^{2}a$
mà $3(a^{2}+ b^{2} +c^{2})\geq (a+b+c)^{2} =9 (theo bunhiacopxki ) \Rightarrow a^{2}+ b^{2}+ c^{2}\geq 3.$
TT cm $a^{4}+ b^{4}+ c^{4}\geq \frac{(a^{2} +b^{2} +c^{2})^{2}}{3} $
$\Rightarrow P \geq \frac{7(a^{2}+ b^{2}+c^{2} )^{2}}{3} +\frac{ab+bc+ca}{a^{2}+b^{2} +c^{2}}$
$=\frac{7(a^{2}+b^{2}+c^{2})^{2}}{3}+\frac{9-(a^{2}+b^{2}+c^{2})}{2(a^{2}+b^{2}+c^{2})}$
đặt t=$a^{2} +b^{2}+c^{2} ,t\geq 3$
khi đó $P\geq \frac{7t^{2}}{3}+\frac{9t}{2}-\frac{1}{2}$ =$\frac{t^{2}}{12} +\frac{9}{4t}+ \frac{9}{4t}+\frac{9t^{2}}{4}-\frac{1}{2}$
$\geq 3\sqrt[3]{\frac{t^{2}}{12}\frac{9}{4t}\frac{9}{4t}} +\frac{9.9}{4}-\frac{1}{2}=22$
dấu "="$\Leftrightarrow a=b=c=1$