How can I prove this $\frac{|xy|}{2x^2+y^2}\le1 $ ? I was thinking about considering the left term a function and maybe show that 1 is the extreme point
x,y can be any real number but not 0
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Can $x,y$ be any real number? It might also be worth mentioning that $x$ and $y$ cannot both be zero at the same time. – graydad Jun 09 '15 at 19:04
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yes , x,y can be any real number but not 0 – user247001 Jun 09 '15 at 19:08
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we have $$2x^2 + y^2 \ge 2\sqrt 2 |x||y| \Rightarrow \frac{|xy|}{2x^2 + y^2} \le \frac 1 {2\sqrt 2} < 1.$$
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I don't understand the downvote. This provides a stronger statement, namely (for $x^2+y^2\neq 0$) that $\frac{|xy|}{2x^2+y^2}\le \frac{1}{2\sqrt{2}}$, with equality iff $\sqrt{2}|x|=|y|$. – user26486 Jun 09 '15 at 19:29
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For $x>0$ and $y>0$ or $x<0$ and $y<0$ $$\frac{xy}{2x^2+y^2}$$ $$\frac{1}{\frac{2x}{y}+\frac{y}{x}}$$ here the denominator always is grater than 1
For $x>0$ and $y<0$ or $x<0$ and $y>0$ $$\frac{-xy}{2x^2+y^2}$$ here the value always is negetive
E.H.E
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