Cm bđt

C

congchuaanhsang

$x>1$ \Leftrightarrow $x^2>1$ \Leftrightarrow $x^1-1>0$

Có $VT=\dfrac{(x^2-1)^2+2x^2}{x(x^2-1)}=\dfrac{x^2-1}{x}+\dfrac{2x}{x^2-1}$
\geq $2\sqrt{2}$ (Cauchy)
 
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