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Can anybody give me a hint on how to bound $$\frac{3xyz}{x^2+y^2+z^2}$$

I'm trying to prove that it converges to zero. Thanks!

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    $\begingroup$ This is the second question that you've asked recently that seems to have the same idea. Perhaps it might be good to understand one before asking the next. $\endgroup$ – robjohn Mar 24 '14 at 15:57
  • $\begingroup$ Yes, I thought of using the AMGM inequality like the question that I've already asked, I just wanted to find another bound. For some reason I don't like to take the limit of an square root. Is not that I didn't understood, I just was looking for other methods to do it. $\endgroup$ – Lessa121 Mar 24 '14 at 16:00
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Hint: Note that $f(x,y,z)\to0$ if and only if $\lvert f(x,y,z)\rvert\to0$. Now, $$ \left\lvert\frac{3xyz}{x^2+y^2+z^2}\right\rvert=\frac{3\lvert x\rvert\,\lvert y\rvert\,\lvert z\rvert}{x^2+y^2+z^2}. $$ Now, note that $$ \lvert x\rvert=\sqrt{x^2}\leq\sqrt{x^2+y^2+z^2}, $$ and similarly $\lvert y\rvert,\lvert z\rvert\leq\sqrt{x^2+y^2+z^2}$.

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  • $\begingroup$ Yes! I did that, and then I got stuck... could you give me a tiny tiny hint on how to keep going? $\endgroup$ – Lessa121 Mar 24 '14 at 15:50
  • $\begingroup$ @LunaSage There really isn't much more to do. Are you trying to do this based on $\epsilon-\delta$ proof, or just squeeze theorem? $\endgroup$ – Nick Peterson Mar 24 '14 at 15:56
  • $\begingroup$ Squeeze theorem $\endgroup$ – Lessa121 Mar 24 '14 at 15:57
  • $\begingroup$ Should I use it on the delta-epsilon proof? $\endgroup$ – Lessa121 Mar 24 '14 at 15:58
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    $\begingroup$ @LunaSage Well, in this case, they're basically the same. Squeeze theorem is fine. The inequality above gives you $0\leq\lvert f(x,y,z)\rvert\leq\sqrt{x^2+y^2+z^2}$, where $f$ is your function. But obviously $0\to0$ as $(x,y,z)\to(0,0,0)$; further, $x^2+y^2+z^2\to0$ (it is a polynomial), so that $\sqrt{x^2+y^2+z^2}\to0$ as well. $\endgroup$ – Nick Peterson Mar 24 '14 at 15:59
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Hint: You can use spherical coordinates:

$$ x=r \, \sin\theta \, \cos\varphi \\ y=r \, \sin\theta \, \sin\varphi \\ z=r \, \cos\theta , $$

where $r=\sqrt{x^2+y^2+z^2}$, $0\leq \theta \leq \pi$, and $0\leq \phi \leq 2\pi$.

Note: consider taking the limit as $r\to 0$.

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Hint: The AM-GM gives $$ \left(x^2y^2z^2\right)^{1/3}\le\frac13\left(x^2+y^2+z^2\right) $$ then $$ \left|\frac{3xyz}{x^2+y^2+z^2}\right|\le\left|\frac{xyz}{\left(x^2y^2z^2\right)^{1/3}}\right|=|xyz|^{1/3} $$


Alternate Approach

Easily, we have $$ \begin{align} x^2&\le x^2+y^2+z^2\\ y^2&\le x^2+y^2+z^2\\ z^2&\le x^2+y^2+z^2 \end{align} $$ Multiplying and taking the square root yields $$ |xyz|\le\left(x^2+y^2+z^2\right)^{3/2} $$ Therefore, $$ \frac{3|xyz|}{x^2+y^2+z^2}\le3\left(x^2+y^2+z^2\right)^{1/2} $$

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Hint: $$\frac{x^2+y^2+z^2}{3} \geq |xyz|^{2/3}$$

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$$ \left|\frac{3xyz}{x^2+y^2+z^2}\right|=\frac{|3xyz|}{\|(x,y,z)\|^2}\le 3\frac{\|(x,y,z)\|^3}{\|(x,y,z)\|^2}=3\|(x,y,z)\| $$

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