cuduckien
BĐT cần cm tương đương với :
[math]3-\sum(\dfrac{x+y}{2})^2+2\sum(\sqrt{1-(\dfrac{x+y}{2}})^2.\sqrt{1-(\dfrac{y+z}{2}})^2 ) \geq6[/math][math]\Leftrightarrow \sum(\sqrt{1-(\dfrac{x+y}{2})^2}.\sqrt{1-(\dfrac{y+z}{2}})^2) \geq \dfrac{7}{4}+\dfrac{xy+yz+zx}{4}[/math]Thật vậy , theo BĐT Cauchy-Schwarz :
[math]\sum(\sqrt{1-(\dfrac{x+y}{2})^2}.\sqrt{1-(\dfrac{y+z}{2}})^2)=\sum (\sqrt{\dfrac{(x-y)^2}{4}+\dfrac{x^2+1}{2}}.\sqrt{\dfrac{(z-y)^2}{4}+\dfrac{z^2+1}{2}})[/math][math]\geq \sum(\dfrac{(x-y)(z-y)}{4}+\dfrac{\sqrt{(x^2+1)(z^2+1)}}{2})[/math][math]\geq \sum(\dfrac{(x-y)(z-y)}{4}+\dfrac{xz+1}{2})=\dfrac{7}{4}+\dfrac{xy+yz+zx}{4}[/math]Dấu bằng xảy ra khi : [math]x=y=z=\dfrac{1}{\sqrt{3}}[/math]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
