3
$\begingroup$

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, $$ x^2+y^2>1. $$

$\endgroup$
  • 1
    $\begingroup$ 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)? $\endgroup$ – TonyK Sep 28 '14 at 16:24
  • $\begingroup$ 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... $\endgroup$ – jm324354 Sep 28 '14 at 16:30
  • $\begingroup$ But they have not referred to your proof at all, except to recommend a couple of irrelevant changes of wording. $\endgroup$ – TonyK Sep 28 '14 at 17:04
  • $\begingroup$ How about now? I've re-written it, let me know what you think. $\endgroup$ – jm324354 Sep 28 '14 at 17:05
  • $\begingroup$ 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). $\endgroup$ – TonyK Sep 28 '14 at 17:19
17
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I totally get how you went to $1<2xy$, but why compare it as $1<2xy\leq{x}^2+y^2$ ? $\endgroup$ – jm324354 Sep 28 '14 at 15:46
  • $\begingroup$ Write $0\le x^2-2xy+y^2$ as $2xy\le x^2+y^2$. $\endgroup$ – Dietrich Burde Sep 28 '14 at 15:47
  • $\begingroup$ 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! $\endgroup$ – jm324354 Sep 28 '14 at 15:54
  • $\begingroup$ @bd1251252 If $a\ge b$ and $b>c$, then always $a>c$ (what you wrote sounds pretty confusing). $\endgroup$ – user144248 Sep 28 '14 at 16:01
  • 3
    $\begingroup$ @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 $x^2+y^2>1$. $\endgroup$ – user144248 Sep 28 '14 at 16:21
2
$\begingroup$

Since $x,y\not=0\Rightarrow x^2,y^2\gt 0$, by AM-GM inequality, we have $$x^2+y^2\ge 2\sqrt{x^2y^2}=2|xy|=2xy\gt 2\cdot \frac 12=1.$$

$\endgroup$
  • $\begingroup$ Also a neat answer. I had to look up AM-GM inequality because I had never heard of it before, interesting! $\endgroup$ – jm324354 Sep 28 '14 at 15:59
  • $\begingroup$ @bd1251252: See here. $\endgroup$ – mathlove Sep 28 '14 at 16:01
1
$\begingroup$

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)

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – jm324354 Sep 28 '14 at 16:16
1
$\begingroup$

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 $x^2+y^2≥2xy$.

$\endgroup$
-1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ That doesn't look simpler than what was posted almost 4 years ago. $\endgroup$ – Henrik Jun 23 '18 at 10:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.