Ta có
$2M=\frac{2x}{2x+y}+\frac{2y}{2y+z}+\frac{2z}{2z+x}$
$=\sum (1-\frac{y}{2x+y})=3-(\frac{y}{2x+y}+\frac{z}{2y+z}+\frac{x}{2z+x})$ $(1)$
Áp dụng bdt Cauchy Shwarz
$\sum \frac{y}{2x+y}=\sum \frac{y^2}{2xy+y^2}\geqslant \frac{(x+y+z)^2}{(x+y+z)^2}=1 (2)$
$(1);(2)\Rightarrow 2M\leqslant 2\Rightarrow M\leqslant 1$
Vậy Max $M=1$
Dấu ''='' xảy ra khi nào?
- lahantaithe99 yêu thích