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

Question

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

See also: http://math.stackexchange.com/questions/470221/prove-the-inequality-xy-leq-frac12x2y2 — Martin Sleziak, Jan 30 '16 at 10:15

score 20 · Answer 1 · answered Mar 04 '13 at 07:22

20

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.

answered Mar 04 '13 at 07:22

André Nicolas

score 11 · Answer 2 · edited Apr 09 '13 at 09:34

11

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

edited Apr 09 '13 at 09:34

Siméon

answered Mar 04 '13 at 07:24

Adi Dani

1

If you're taking this approach, you should say something about when $x-y<0$. – Clayton Mar 04 '13 at 07:25
3

@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. – hjpotter92 Mar 04 '13 at 07:37
then WLOG we can say..maybe you should mention it to the OP.. – Belgi Mar 04 '13 at 11:08
1

btw, you don't use that x > y, you actually use x != y – RiaD Mar 04 '13 at 12:24

score 2 · Answer 3 · answered Mar 04 '13 at 12:21

2

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.

answered Mar 04 '13 at 12:21

galmeida

3 Answers3