Đặt \(\left(a;b;c\right)=\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\Rightarrow xyz=1\)
\(P=\frac{x^3yz}{y+z}+\frac{xy^3z}{x+z}+\frac{xyz^3}{x+y}=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
\(P\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\ge\frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)