\(x^3\)hay \(x^2\)
1/ \(P=\frac{1}{a\left(1+b\right)}+\frac{1}{b\left(1+c\right)}+\frac{1}{c\left(1+a\right)}\)
Đặt \(abc=k^3\)
Khi đó luôn tồn tại \(x;y;z\) dương sao cho \(\left\{{}\begin{matrix}a=\frac{ky}{x}\\b=\frac{kz}{y}\\c=\frac{kx}{z}\end{matrix}\right.\)
\(\Rightarrow P=\frac{1}{\frac{ky}{x}\left(1+\frac{kz}{y}\right)}+\frac{1}{\frac{kz}{y}\left(1+\frac{kx}{z}\right)}+\frac{1}{\frac{kx}{z}\left(1+\frac{ky}{x}\right)}=\frac{x}{k\left(y+kz\right)}+\frac{y}{k\left(z+kx\right)}+\frac{z}{k\left(x+ky\right)}\)
\(\Rightarrow P=\frac{1}{k}\left(\frac{x^2}{xy+kzx}+\frac{y^2}{yz+kxy}+\frac{z^2}{zx+kyz}\right)\ge\frac{\left(x+y+z\right)^2}{k\left(k+1\right)\left(xy+yz+zx\right)}\)
\(\Rightarrow P\ge\frac{3\left(xy+yz+zx\right)}{k\left(k+1\right)\left(xy+yz+zx\right)}=\frac{3}{k\left(k+1\right)}\)
Mặt khác ta luôn có: \(k^3+1\ge k\left(k+1\right)\) với mọi k dương
Thật vậy, \(\Leftrightarrow\left(k+1\right)\left(k^2-k+1\right)\ge k\left(k+1\right)\)
\(\Leftrightarrow k^2-k+1\ge k\)
\(\Leftrightarrow\left(k-1\right)^2\ge0\) (luôn đúng)
\(\Rightarrow P\ge\frac{3}{k\left(k+1\right)}\ge\frac{3}{1+k^3}=\frac{3}{1+abc}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
