Đặt \(a=\dfrac{kx}{y};b=\dfrac{ky}{z};c=\dfrac{kz}{x}\Rightarrow abc=k^3\)
Ta có: \(BDT\Leftrightarrow\dfrac{yz}{kx\left(ky+z\right)}+\dfrac{xz}{ky\left(kz+x\right)}+\dfrac{xy}{kz\left(kx+y\right)}\ge\dfrac{3}{1+k^3}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT=\dfrac{y^2z^2}{kxyz\left(ky+z\right)}+\dfrac{x^2z^2}{kxyz\left(kz+x\right)}+\dfrac{x^2y^2}{kxyz\left(kx+y\right)}\)
\(\ge\dfrac{\left(xy+yz+xz\right)^2}{xyz\left(x+y+z\right)k\left(k+1\right)}\ge\dfrac{3xyz\left(x+y+z\right)}{xyz\left(x+y+z\right)k\left(k+1\right)}=\dfrac{3}{k\left(k+1\right)}\)
Cần chứng minh \(\dfrac{3}{k\left(k+1\right)}\ge\dfrac{3}{1+k^3}\)
\(\Leftrightarrow\dfrac{3\left(k-1\right)^2}{k\left(k+1\right)\left(k^2-k+1\right)}\ge0\) (luôn đúng)
