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I have an inequality to prove and I can't get a hold of it... I hope someone can help with it or point me in the right direction.

$x,y\in\mathbb{R},\quad \epsilon\in\mathbb{R}:\epsilon\not=0$

$$ 2xy\leq \epsilon^2 x^2 + y^2/\epsilon^2 $$

I've tried to transform it based on previous exercises, but I'm not sure how to start it... Any help would be appreciated!

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  • $\begingroup$ Can you do the case $\epsilon=1$? Then try $\tilde x=x \epsilon$ and $\tilde y=y/\epsilon$. $\endgroup$ – Jochen Nov 20 '14 at 16:04
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multiply both sides by $\epsilon^2$ (which preserves the direction of the inequality because it's positive) and move everything on one side of the inequality. You will be able to see that now the expression is $$0 ≤ \text{the square of something}$$.

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  • $\begingroup$ I've tried this, I first moved $2xy$ to the other side, multiplied by $\epsilon^2$, but I can't see the square in $\epsilon^4 x^2 + y^2 -2xy\epsilon^2$ $\endgroup$ – Petra Nov 20 '14 at 16:08
  • $\begingroup$ $\epsilon^4x^2 - 2\epsilon^2xy + y^2$ has a better shape. Can you see it now? $\endgroup$ – rewritten Nov 20 '14 at 16:10
  • $\begingroup$ $(\epsilon^2x-y)^2$? $\endgroup$ – Petra Nov 20 '14 at 16:12
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$$ (\epsilon x - y/\epsilon)^2\geq 0 $$

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  • $\begingroup$ could you detail this? $\endgroup$ – Petra Nov 20 '14 at 16:07
  • $\begingroup$ @PetraD. just open the brackets, and move the $xy$ term to another side - you will get your inequality. $\endgroup$ – Ilya Nov 20 '14 at 16:16

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