Question

Giúp mình với ạ

Answer 1

\(\left(x^2+1\right)+\left(y^2+1\right)+\left(z^2+1\right)+2\left(x^2+y^2+z^2\right)\ge2x+2y+2z+2\left(xy+yz+zx\right)\)

\(\Leftrightarrow3\left(x^2+y^2+z^2\right)+3\ge12\)

\(\Rightarrow x^2+y^2+z^2\ge3\)

Ta có:

\(\dfrac{x^3}{y}+\dfrac{y^3}{z}+\dfrac{z^3}{x}=\dfrac{x^4}{xy}+\dfrac{y^4}{yz}+\dfrac{z^4}{zx}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{xy+yz+zx}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2}\ge3\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=1\)