\(P=\frac{a^3+1+1+6}{a^3\left(b+c\right)}+\frac{b^3+1+1+6}{b^3\left(a+c\right)}+\frac{c^3+1+1+6}{c^3\left(a+b\right)}\)
\(P\ge\frac{3a+6}{a^3\left(b+c\right)}+\frac{3b+6}{b^3\left(a+c\right)}+\frac{3c+6}{c^3\left(a+b\right)}\)
Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xyz=1\)
\(P\ge\frac{\left(3x^2+6x^3\right)yz}{y+z}+\frac{\left(3y^2+6y^3\right)xz}{x+z}+\frac{\left(3z^2+6z^3\right)xy}{x+y}=\frac{3x+6x^2}{y+z}+\frac{3y+6y^2}{x+z}+\frac{3z+6z^2}{x+y}\)
\(P\ge3\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)+6\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)\)
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