Question

Given x>0 , y>0 and x y , compare the size of \dfrac{x^2}{y^2} +\dfrac{y^2}{x^2} and \dfrac{x}{y}+\dfrac{y}{x}

Steven Zheng Steven Zheng posted 1 week ago


Answer

Define t=\dfrac{x}{y}+\dfrac{y}{x}

\because x y

\therefore t>2

\dfrac{x^2}{y^2} +\dfrac{y^2}{x^2}-( \dfrac{x}{y}+\dfrac{y}{x})

= t^2-t-2

=(t-2)(t+1)

>0

\therefore \dfrac{x^2}{y^2} +\dfrac{y^2}{x^2} > \dfrac{x}{y}+\dfrac{y}{x}

Steven Zheng Steven Zheng posted 1 week ago

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