Đặt [tex]p=a+b+c,q=ab+bc+ca,r=abc=1\Rightarrow P=\frac{q+4p+12}{r+2q+4p+8}=\frac{q+4p+12}{2q+4p+9}[/tex]
Vì [tex]q=ab+bc+ca\geq 3\sqrt[3]{a^2b^2c^2}=1\Rightarrow P=\frac{q+4p+12}{2q+4p+9}\leq \frac{q+4p+12}{q+3+4p+9}=\frac{q+4p+12}{q+4p+12}=1[/tex]
Dấu "=" xảy ra khi a = b = c = 1.