Cho $x,y,z\in \left ( 0,1 \right ]$ .CMR
$\frac{1}{1+xy}+\frac{1}{1+yz}+\frac{1}{1+zx}\leq \frac{5}{x+y+z}$
Cho $x,y,z\in \left ( 0,1 \right ]$ .CMR
$\frac{1}{1+xy}+\frac{1}{1+yz}+\frac{1}{1+zx}\leq \frac{5}{x+y+z}$
Cho $x,y,z\in \left ( 0,1 \right ]$ .CMR
$\frac{1}{1+xy}+\frac{1}{1+yz}+\frac{1}{1+zx}\leq \frac{5}{x+y+z}$
Ta có:$(1-x)(1-y)\geq 0= > xy+1\geq x+y= > \frac{1}{xy+1}\leq \frac{1}{x+y}$.Tương tự $\frac{1}{yz+1}\leq \frac{1}{y+z},\frac{1}{xz+1}\leq \frac{1}{x+z}$
Cộng theo vế $= > \sum \frac{1}{xy+1}\leq \sum \frac{1}{x+y}$
Do đó ta cần CM:$\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\leq \frac{5}{x+y+z}< = > \left [ (x+y)+(x+z)+(y+z) \right ](\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z})\leq 10$
Đặt $x+y=a,y+z=b,x+z=c= > 0\leq a,b,c\leq 2$
BDT$< = > (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\leq 10$.Đây chính là bdt quen thuộc
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