Prove that for any two real numbers $x,y$ with $x\neq 0$

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Prove that for any two real numbers $x,y$ with $x\neq 0$

$$2y \leq \frac{y^2}{x^2}+x^2$$

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Hint:

Explore the following expression $\left( x-\frac{y}{x}\right)^2$

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By AM-GM we obtain: $$\frac{y^2}{x^2}+x^2\geq2\sqrt{\frac{y^2}{x^2}\cdot x^2}=2|y|\geq2y.$$ Done!