Question

cho x,y,z >0 thỏa mãn :xyz=1 . c/m : \(\dfrac{x^4y}{x^2+1}+\dfrac{y^4z}{y^2+1}+\dfrac{z^4x}{z^2+1}\ge\dfrac{3}{2}\)

Answer 1

\(\sum\dfrac{x^4y}{x^2+1}=\sum\dfrac{x^3.\dfrac{1}{z}}{x^2+xyz}=\sum\dfrac{x^2}{z\left(x+yz\right)}=\sum\dfrac{x^2}{xz+1}\)

Áp dụng bất đẳng thức cauchy-schwarz:

\(Vt=\sum\dfrac{x^2}{xz+1}\ge\dfrac{\left(x+y+z\right)^2}{xy+yz+xz+3}\)

mà theo AM-GM: \(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}=3\)

hay \(3\le xy+yz+xz\)

do đó \(VT\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\dfrac{3\left(xy+yz+zx\right)}{2\left(xy+yz+xz\right)}=\dfrac{3}{2}\)

Dấu = xảy ra khi x=y=z=1

P/s: Câu này khoai