[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
$\dfrac{\sqrt{x-1}}{x}$=$\dfrac{\sqrt{1(x-1}}{x}$\leq$\dfrac{1+x-1}{2x}$=$\dfrac{1}{2}$
$\dfrac{\sqrt{y-1}}{y}$=$\dfrac{\sqrt{1(y-1)}}{y}$\leq$\dfrac{1+y-1}{y}$=$\dfrac{1}{2}$
\Rightarrow$\dfrac{\sqrt{x-1}}{x}$+$\dfrac{\sqrt{y-1}}{y}$\leq1
Dấu "=" xảy ra \Leftrightarrow x=y=2
 
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