Question

Cho \(x\ge y\ge z\ge0\). Chứng minh BĐT sau

a/ \(xy^3+yz^3+zx^3\ge xz^3+zy^3+yx^3\)

b/ \(\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\ge\dfrac{x^2z}{y}+\dfrac{y^2x}{z}+\dfrac{z^2y}{x}\)

Answer 1

a) BĐT \(\Leftrightarrow\left(x-y\right)\left(y-z\right)\left(z-x\right)\left(x+y+z\right)\ge0\)

suy ra sai đề

b) BĐT \(\Leftrightarrow\dfrac{\left(x-y\right)\left(y-z\right)\left(x-z\right)\left(xy+yz+xz\right)}{xyz}\ge0\) ( đúng vì \(x\ge y\ge z>0\))