Đặt $x=\frac{a(b+c)}{a^2+bc}$ $y=...$, $z=...$ $\Rightarrow P=x+y+z$
Xét $A=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\Sigma \frac{a^2+bc}{a(b+c)}$
$\Rightarrow A+3=\Sigma (\frac{a^2+bc}{a(b+c)}+1)=\frac{(a+b)(a+c)}{a(b+c)}$
A/d BĐT C.S cho 3 số dg, ta đc:
$A\geq 3\sqrt[3]{\frac{(a+b)^2.(b+c)^2.(c+a)^2}{abc(a+b)(b+c)(c+a)}}=3\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{abc}}\geq 3\sqrt[3]{2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}/abc}=3\sqrt[3]{\frac{8abc}{abc}}=6$
$\rightarrow A\geq 6-3=3$
Dấu đẳng thức xảy ra khi $a=b=c$
Dễ c/m đc: $\Sigma \frac{1}{x}\geq \frac{9}{x+y+z}$ hay $A.P\geq 9$
mà$A\geq3$ suy ra $P\leq 3$
$\Rightarrow ....$