Question

Cho các số thực dương x,y,z thỏa mãn x+y+z=2

CMR: \(\dfrac{x^2}{x+y}\)+\(\dfrac{y^2}{y+z}\)+\(\dfrac{z^2}{z+x}\)\(\ge\)1

Answer 1

Theo BĐT Cô-si dưới dạng engel ta có :

\(\dfrac{x^2}{x+y}+\dfrac{y^2}{y+z}+\dfrac{z^2}{z+x}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{4}{4}=1\)

Dấu \("="\) xảy ra khi \(x=y=z=\dfrac{2}{3}\)

Answer 2

Cách khác :

\(\dfrac{x^2}{x+y}+\dfrac{x+y}{4}\ge2.\sqrt{\dfrac{x^2}{x+y}.\dfrac{x+y}{4}}=x\\ \dfrac{y^2}{y+z}+\dfrac{y+z}{4}\ge y\\ \dfrac{z^2}{x+z}+\dfrac{x+z}{4}\ge z\\ \Rightarrow\dfrac{x^2}{x+y}+\dfrac{y^2}{y+z}+\dfrac{z^2}{x+z}+\dfrac{x+y+z}{2}\ge x+y+z\\ \Rightarrow\dfrac{x^2}{x+y}+\dfrac{y^2}{y+z}+\dfrac{z^2}{x+z}\ge2-1=1\)