[tex]\sqrt{x-2}+\sqrt{y+2009}+\sqrt{z-2010}=\frac{1}{2}[/tex] (x+y+z)
ĐKXĐ: [tex]x\geq 2;y\geq -2009;z\geq 2010[/tex]
Áp dụng BĐT Cauchy ta có:
[tex]\sqrt{x-2}=\sqrt{(x-2).1}\leq \frac{x-2+1}{2}=\frac{x-1}{2}[/tex]
[tex]\sqrt{y+2009}=\sqrt{(y+2009).1}\leq \frac{y+2009+1}{2}= \frac{y+2010}{2}[/tex]
[tex]\sqrt{z-2010}=\sqrt{(z-2010).1}\leq \frac{z-2010+1}{2}= \frac{z-2009}{2}[/tex]
Cộng vế với vế các BĐT trên:
[tex]\sqrt{x-2}+\sqrt{y+2009}+\sqrt{z-2010}\leq \frac{x+y+z}{2}=\frac{1}{2}.(x+y+z)[/tex]
Dấu "=" xảy ra <=> x-2=1; y+2009=1; z-2010=1
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