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Show that $2 xy < x^2 + y^2$ for $x$ is not equal to $y$.
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.
If $x\neq y$ then without loss of generality we can assume that $x>y$ $x-y>0\Rightarrow (x-y)^2>0\Rightarrow x^2-2xy+y^2>0\Rightarrow x^2+y^2>2xy$
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.
Required, but never shown