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Show that $2 xy < x^2 + y^2$ for $x$ is not equal to $y$.

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Note that $(x-y)^2 \ge 0$ for all real numbers $x$ and $y$, with equality only when $x=y$. Now expand $(x-y)^2$.

A surprisingly large number of useful inequalities follow from the simple observation that any square is non-negative.

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If $x\neq y$ then without loss of generality we can assume that $x>y$ $x-y>0\Rightarrow (x-y)^2>0\Rightarrow x^2-2xy+y^2>0\Rightarrow x^2+y^2>2xy$

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    $\begingroup$ If you're taking this approach, you should say something about when $x-y<0$. $\endgroup$ – Clayton Mar 4 '13 at 7:25
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    $\begingroup$ @Clayton I don't think so. $x$ and $y$ are variables, they can always change their values to satisfy the condition defined/explained/derived above. $\endgroup$ – hjpotter92 Mar 4 '13 at 7:37
  • $\begingroup$ then WLOG we can say..maybe you should mention it to the OP.. $\endgroup$ – Belgi Mar 4 '13 at 11:08
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    $\begingroup$ btw, you don't use that x > y, you actually use x != y $\endgroup$ – RiaD Mar 4 '13 at 12:24
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Let $$y=x+a$$ $$a \neq 0$$ then $$2xy=2x(x+a)=2x^2+2ax<2x^2+2ax+a^2$$

see that the difference is exactly 'a' squared.

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