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Let $x$ and $y$ be real numbers. If $x\cdot{y}>\frac{1}{2}$, then $x^2+y^2>1$.

Proof: We will prove with the direct method. Let $x$ and $y$ be real numbers. Since $$ x\cdot{y}>\frac{1}{2} $$ it follows that $$ 2xy>1,$$ which means $$x^2+y^2 \geq 2xy.$$ Therefore, $$ x^2+y^2>1. $$

Christopher King
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bjd2385
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    I can't follow this at all. $a$ means $x$ and $b$ means $y$! And everything is backwards! How does $$a^2+\left(\frac{1}{2a}\right)^2 > 1$$ follow from $a \cdot b > \dfrac{1}{2}$? What's this "which thus allows", where you should have "which thus proves" (but it doesn't)? – TonyK Sep 28 '14 at 16:24
  • Yeah, it's all wrong, I know that. That's why I was asking how to make it better, and below you'll see that a few people obviously answered... – bjd2385 Sep 28 '14 at 16:30
  • But they have not referred to your proof at all, except to recommend a couple of irrelevant changes of wording. – TonyK Sep 28 '14 at 17:04
  • How about now? I've re-written it, let me know what you think. – bjd2385 Sep 28 '14 at 17:05
  • That's much better! But your "which means" is wrong. $x^2+y^2\ge 2xy$ is not a consequence of $2xy > 1$, it is true in its own right (and you should briefly explain why). – TonyK Sep 28 '14 at 17:19

5 Answers5

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Write $0\le (x-y)^2=x^2-2xy+y^2$, hence $1<2xy\le x^2+y^2$.

Dietrich Burde
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  • I totally get how you went to $1<2xy$, but why compare it as $1<2xy\leq{x}^2+y^2$ ? – bjd2385 Sep 28 '14 at 15:46
  • Write $0\le x^2-2xy+y^2$ as $2xy\le x^2+y^2$. – Dietrich Burde Sep 28 '14 at 15:47
  • Oh I see, so given that $1<2xy$, we may say also that $x^2+y^2>1$ can be rewritten as $x^2+y^2>2xy$, substituting $2xy$ for $1$, which also means that, as you've already written, $x^2+y^2>2xy>1$. Thanks! – bjd2385 Sep 28 '14 at 15:54
  • @bd1251252 If $a\ge b$ and $b>c$, then always $a>c$ (what you wrote sounds pretty confusing). – user144248 Sep 28 '14 at 16:01
  • Out of curiosity, is it appropriate to say that, given $xy>\frac{1}{2}$, $x^2+y^2>1\longrightarrow x^2+y^2>2xy>1$, or should I use biconditional to say that $x^2+y^2>1 \longleftrightarrow x^2+y^2>2xy>1$ ? – bjd2385 Sep 28 '14 at 16:04
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    @bd1251252 Your logic is entirely messed up. It's not that $xy>1/2$ and $x^2+y^2>1$ imply $x^2+y^2>2xy>1$ (recall that you want to prove that $x^2+y^2>1$ so you can't put this in the assumptions). You start with $x^2+y^2\ge 2xy$ and $2xy>1$ and from those two things you conclude that $x^2+y^2>1$. – user144248 Sep 28 '14 at 16:21
  • Okay, that makes sense. I figure it's gonna burn for a while until I figure things out, and how to draw conclusions from the statements. – bjd2385 Sep 28 '14 at 16:37
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Since $x,y\not=0\Rightarrow x^2,y^2\gt 0$, by AM-GM inequality, we have $$x^2+y^2\ge 2\sqrt{x^2y^2}=2|xy|=2xy\gt 2\cdot \frac 12=1.$$

mathlove
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I'm probably going to be a little grammatically picky, but I personally think proper grammar is good in a proof, (and an analysis lecturer drilled it into our class to use 'proper English sentences' where the maths would read as part of the sentence).

  • I would remove the comma after the first line of mathematical notation say something like 'we have/get', instead of 'therefore'.

  • 'If we substitute...' (no and)

Dom
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  • Thanks! I'm just learning how to write basic proofs and things like this from people who have done more than I are really great. – bjd2385 Sep 28 '14 at 16:16
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The only problem is the "which means".

$2xy>1$ doesn't imply $x^2+y^2≥2xy$ in any way.

$x^2+y^2≥2xy$ is true because $x^2-2xy+y^2 = (x-y)^2 ≥ 0$, which means $x^2+y^2≥2xy$.

gnasher729
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There is a much simpler proof.

Since $xy > \frac{1}{2}$, $x$ and $y$ must be non zero numbers, which are both positive or negative.

Suppose $x>0$ and $y>0$.

If $x>1$, then also $x^2>1$, and $x^2 +y^2>x^2>1$ is obvious.

Take $0<x\leq 1$ and $y>0$.

Since $0<x\leq 1$, the function $f$ with $f(x)=\sqrt{1-x^2}$ is well defined. Let $g$ be the function with $g(x)=\frac{1}{2x}, \ 0<x\leq 1$.

It can be easily proved that the function h with $h(x)=-x+\sqrt{2}, \ 0<x\leq 1$ has a graph which is a line tangent to the graphs of $f$ and $g$ at the point $(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$.

One can also verify easily that $f$ is a concave function and that $g$ is a convex funtion. Therefore the tangent line $h$ is above the graph of $f$ and below the graph of $g$. So if $0<x\leq 1$,the previous remark implies that $f(x)\leq h(x)\leq g(x)$ and $\sqrt{1-x^2}\leq \frac{1}{2x}$.

Now, the proposition we want to prove is obvious: If $xy > \frac{1}{2}$, then $y>\frac{1}{2x}$, and $\frac{1}{2x} \geq \sqrt{1-x^2}$ from the previous inequality, hence $y> \frac{1}{2x} \geq \sqrt{1-x^2}$. Consequently, $y>\sqrt{1-x^2}$, which implies $y^2>1-x^2$ and therefore $x^2+y^2>1.$

The case $x<0$ and $y<0$ is similar.

kafroulis
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