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

Question

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

Uniteasy Admin · Accepted Answer

Define $$ t=\dfrac{x}{y}+\dfrac{y}{x} $$
$$ \because x $$ ≠ $$ y $$ 
$$ 	herefore t>2 $$
$$ \dfrac{x^2}{y^2} +\dfrac{y^2}{x^2}-( \dfrac{x}{y}+\dfrac{y}{x}) $$  
$$ = t^2-t-2 $$
$$ =(t-2)(t+1) $$
$$ >0 $$
$$ 	herefore  \dfrac{x^2}{y^2} +\dfrac{y^2}{x^2} > \dfrac{x}{y}+\dfrac{y}{x} $$

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