Đề phải là : \(\frac{9}{2\left(a+b+c\right)}\)
Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\left(x;y;z>0\right)\)
\(\Rightarrow\frac{a}{b^2}=\frac{y^2}{x};\frac{b}{c^2}=\frac{z^2}{y};\frac{c}{a^2}=\frac{x^2}{z};xyz=1\)
\(\frac{9}{2\left(a+b+c\right)}=\frac{9}{\frac{2\left(a+b+c\right)}{abc}}\left(abc=1\right)=\frac{9}{2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)}=\frac{9}{2\left(xy+yz+xz\right)}\)
Khi đó , ta có : \(P=\frac{x^2}{z}+\frac{y^2}{x}+\frac{z^2}{y}+\frac{9}{2\left(xy+yz+xz\right)}\)
\(=\frac{x^2}{z}+z+\frac{y^2}{x}+x+\frac{z^2}{y}+y+\frac{9}{2\left(xy+yz+xz\right)}-x-y-z\)
AD BĐT Cauchy , ta có :
\(P\ge2x+2y+2z+\frac{9}{\frac{2\left(x+y+z\right)^2}{3}}-\left(x+y+z\right)=x+y+z+\frac{27}{2\left(x+y+z\right)^2}\)
\(=\frac{x+y+z}{2}+\frac{x+y+z}{2}+\frac{27}{2\left(x+y+z\right)^2}\ge3.\sqrt[3]{\frac{27}{8}}=\frac{9}{2}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)
