Question

CHO x,y,z >0 ,xyz=\(\frac{1}{2}\)

CMR:\(\frac{yz}{x^2\left(y+z\right)}\)+\(\frac{zx}{y^2\left(z+x\right)}\)+\(\frac{xy}{z^2\left(x+y\right)}\) ≥ xy+yz+zx

Answer 1

\(VT=\frac{\left(yz\right)^2}{x^2yz\left(y+z\right)}+\frac{\left(zx\right)^2}{xy^2z\left(z+x\right)}+\frac{\left(xy\right)^2}{xyz^2\left(x+y\right)}\)

\(VT=\frac{2\left(yz\right)^2}{xy+xz}+\frac{2\left(zx\right)^2}{xy+yz}+\frac{2\left(xy\right)^2}{xz+yz}\)

\(VT\ge\frac{2\left(xy+yz+zx\right)^2}{2\left(xy+yz+zx\right)}=xy+yz+zx\)

Dấu "=" xảy ra khi \(x=y=z=\frac{1}{\sqrt[3]{2}}\)