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Lấy $a=\min\{a,b,c\}$, $b=a+u$ và $c=a+v,u,v \ge 0$.
Ta có , $P-\frac{23}{3}=\frac{14(a^2+b^2+c^2)}{(a+b+c)^2}+\frac{(ab+bc+ca)(a+b+c)}{a^2b+b^2c+c^2a}-\frac{23}{3}=$
$=\frac{1}{3(a+b+c)^2(a^2b+b^2c+c^2a)}(57(u^2-uv+v^2)a^3+$
$+3(22u^3-18u^2v+9uv^2+22v^3)a^2+(25u^4-11u^3v-18u^2v^2+43uv^3+25v^4)a+$
$+(22u^3-37u^2v+28uv^2+2v^3)uv)\geq0$
Vậy $\min P=\frac{23}{3}$ $\Leftrightarrow a=b=c=\frac{1}{3}$.
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