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This inequality is wrong - see the accepted answer (it appears there is no general inequality for these two expressions).

On the left we have harmonic mean of pairwise geometric means, which obeys:

$$\sqrt[3]{xyz} \geq \color{blue}{ \frac{3\sqrt{xyz}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}} \geq 2 \sqrt[3]{\frac{x^2y^2z^2}{(x+y)(y+z)(z+x)}} \geq \frac{3xyz}{xy+yz+zx}$$

On the right we have arithmetic means of pairwise harmonic means, obeying the same inequality (the first inequality here may not be true as well, waiting for proof in another question):

$$\sqrt[3]{xyz} \geq \color{blue}{ \frac{2}{3} \left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)} \geq 2 \sqrt[3]{\frac{x^2y^2z^2}{(x+y)(y+z)(z+x)}} \geq \frac{3xyz}{xy+yz+zx}$$

According to Wolfram Alpha the following is true:

$$\frac{3\sqrt{xyz}}{\sqrt{x}+\sqrt{y}+\sqrt{z}} \leq \frac{2}{3} \left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)$$

But I have not been able to prove it so far.


It may help to transform the RHS in the following way:

$$\frac{2}{3} \left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)=\frac{2}{3} \frac{xyz(x+y+z)+(xy+yz+zx)^2}{(x+y+z)(xy+yz+zx)-xyz}$$

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  • $\begingroup$ I'm voting to close this question as off-topic because the original question is an unmotivated inequality that is not correct. In this case I think there is little long-term benefit to archiving it on this site, because there is no reason to expect anyone else to look for this particular inequality. $\endgroup$ – Carl Mummert Mar 31 '18 at 19:34
  • $\begingroup$ @CarlMummert, very well $\endgroup$ – Yuriy S Mar 31 '18 at 19:36
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Your inequality is wrong! For $y=z=1$ and $x\rightarrow+\infty$ we obtain:

$3\leq\frac{2}{3}\left(1+\frac{1}{2}+1\right)$, which is wrong.

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  • 1
    $\begingroup$ You are right. I did not expect this to happen. So far every inequality made of mean values worked $\endgroup$ – Yuriy S Jun 6 '16 at 7:31
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Wrong Hint:

$\frac{3\sqrt{xyz}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)^{-1} \leq \frac{2}{3} $

$\to 2.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \leq \frac{2}{3}.\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3\sqrt{xyz}}=\frac{2}{9}.\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}} \right) $

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  • $\begingroup$ How did you get from $$\left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)^{-1}$$ to $$2 \left( \frac{1}{x}+ \frac{1}{y}+ \frac{1}{z} \right)$$? $\endgroup$ – Yuriy S Jun 6 '16 at 7:05
  • $\begingroup$ yes you right ı did mistake $\endgroup$ – user2312512851 Jun 6 '16 at 7:25

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