[Toán 9]Cm bất đẳng thức

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toankyniem

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congchuaanhsang

Áp dụng BĐT Cauchy cho 2 số dương ta có
x−1x\dfrac{\sqrt{x-1}}{x}=1(x−1x\dfrac{\sqrt{1(x-1}}{x}\leq1+x−12x\dfrac{1+x-1}{2x}=12\dfrac{1}{2}
y−1y\dfrac{\sqrt{y-1}}{y}=1(y−1)y\dfrac{\sqrt{1(y-1)}}{y}\leq1+y−1y\dfrac{1+y-1}{y}=12\dfrac{1}{2}
\Rightarrowx−1x\dfrac{\sqrt{x-1}}{x}+y−1y\dfrac{\sqrt{y-1}}{y}\leq1
Dấu "=" xảy ra \Leftrightarrow x=y=2
 
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