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prove: if $x,y\in\mathbb{R}$ then $2xy\le x^2+y^2$

if one or both of $x,y$ is $0$ or one is negative, the result follows. if both are negative the negatives cancel and it is the same as if they were both positive. both positive or both negative:

$2xy\le x^2+y^2\Leftrightarrow 2y\le{x+{y^2\over x}}\Leftrightarrow 2\le {x\over y}+{y\over x}$.$\space\space$if $x=y$ the result follows with ${x\over y}+{y\over x}=2$. if $x\gt y\space$ then $x=y+i,i\in\mathbb{R_+} \space$ and $\space{x\over y}+{y\over x}={{y+i}\over y}+{y\over {y+i}}={y\over y}+{i\over y}+{y\over {y+i}}\gt {y\over y}+{i\over {y+i}}+{y\over {y+i}}$ $=1+{{i+y}\over{i+y}}=2$.$\space\space$similar argument can be used for $y\gt x$.

Is this proof valid for real numbers? especially irrational numbers. thanks.

Martin Sleziak
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miniparser
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  • I think your proof would work (TBH I didn't read it past $2 \le x/y + y/x$; it's a known fact $2 \le x + 1/x$ for positive $x$ so I figured you were rediscovering that.) However a far EASIER proof is $2xy \le x^2 + y^2 \iff x^2 -2xy + y^2 \ge 0 \iff (x - y)^2 \ge 0$ which is, of course, true. – fleablood Jan 17 '17 at 21:23
  • not really a duplicate but the answers do suffice. – miniparser Jan 17 '17 at 21:42
  • Your proof is good. Note: $x > 0$, $x + 1/x \ge 2$. Wolog $x \ge 1$. Let $x + 1/x = x - 1 + 1 + 1/x \ge \frac{x-1}x + \frac 1x + 1 = \frac xx + 1 = 2$. ... same argument... but $x + 1/x \ge 2 \iff x^2 + 1 \ge 2x \iff $x^2 - 2x + 1 \ge 0 \iff (x-1)^2 \ge 0$ might be easier. – fleablood Jan 17 '17 at 22:02
  • i think my proof is better in an obscure way in that it does not rely on any identity etc. such as the binomial. i guess the answers given should have been obvious though – miniparser Jan 18 '17 at 16:43
  • I like your proof. But ... what's wrong with relying on an identity-- especially a very well known one. Even if you don't know the binomial theorem I think $x^2 - 2xy + y^2 = (x - y)^2$ is simply a matter of factoring and can be a single statement. – fleablood Jan 18 '17 at 16:55
  • nothing wrong with relying on an identity and you don't even need to know binomial for the case $n=2$ like you said. it's just good to show different ways of proving something. – miniparser Jan 18 '17 at 17:24
  • See also: Prove the inequality $|xy|\leq\frac{1}{2}(x^2+y^2)$ and Show that $2 xy < x^2 + y^2$ for $x$ is not equal to $y$ – Martin Sleziak Dec 09 '18 at 07:46

2 Answers2

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HINT

$$x^2 + y^2 - 2xy = (x-y)^2$$

gt6989b
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Fundamentally this comes down to proving that $$x^2-2xy+y^2 \ge 0$$ All we have to do is factor the left side to get $$(x-y)^2 \ge 0$$ Note that anything squared will be $\ge 0$

Brevan Ellefsen
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