Chủ đề này có 1 trả lời

[thanhduc991010]

Cho $x+y+z=\frac{3}{2}$

Tìm GTNN: M=$\sum \frac{\sqrt{x^2+xy+y^2}}{4yz+1}$

 

[Hoang Tung 126]

Ta có :$M=\sum \frac{\sqrt{x^2+xy+y^2}}{4yz+1}\geq \sum \frac{\sqrt{\frac{3}{4}(x+y)^2}}{4yz+1}=\frac{\sqrt{3}}{2}.\sum \frac{x+y}{4yz+1}\geq \frac{\sqrt{3}}{2}.\sum \frac{x+y}{(y+z)^2+1}$$M=\sum \frac{\sqrt{x^2+xy+y^2}}{4yz+1}\geq \sum \frac{\sqrt{\frac{3}{4}(x+y)^2}}{4yz+1}=\frac{\sqrt{3}}{2}.\sum \frac{x+y}{4yz+1}\geq \frac{\sqrt{3}}{2}.\sum \frac{x+y}{(y+z)^2+1}$

Đặt $y+z=a,x+z=b,x+y=c$ $= > a+b+c=2(x+y+z)=3$

$= > A\geq\frac{\sqrt{3}}{2}. \sum \frac{c}{a^2+1}=\frac{\sqrt{3}}{2}.\sum \frac{c(a^2+1)-a^2c}{a^2+1}=\frac{\sqrt{3}}{2}.(\sum c-\sum \frac{a^2c}{a^2+1})\geq \frac{\sqrt{3}}{2}.(3-\sum \frac{a^2c}{2a})=\frac{\sqrt{3}}{2}(3-\sum \frac{ac}{2})\geq \frac{\sqrt{3}}{2}(3-\frac{(a+b+c)^2}{6})=\frac{\sqrt{3}}{2}(3-\frac{9}{6})=\frac{3\sqrt{3}}{4}$

 Dấu=  xảy ra tại a=b=c=1 hay $x=y=z=\frac{1}{2}$