Toán [Toán 9] Nâng cao

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phamhuy20011801

$ \sum \dfrac{x}{\sqrt{1+x^2}}=\sum \sqrt{ \dfrac{x^2}{xy+yz+zx+x^2} }= \sum \sqrt{ \dfrac{x}{x+y} . \dfrac{x}{x+z} } \le \sum \dfrac{\dfrac{x}{x+y}+\dfrac{x}{x+z}}{2}=\dfrac{3}{2}$
 
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