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Please help to prove this inequality.

Prove $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 1$ when $(x^2-1)(y^2-1)(z^2-1)=8^3$ and each of $x,y,z$ is greater than 1.

Thanks.

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  • $\begingroup$ This will likely require similar machinery as this other question. You are asking about the inequality on the other side, after a simple change of variables. You have the 'near equality' case of $x \rightarrow 1, y, z \rightarrow \infty$ to take care of. $\endgroup$ – Calvin Lin Jul 19 '13 at 16:26
  • $\begingroup$ Not necessarily. The minimum is reached when the $x,y,z$ are equal here. $\endgroup$ – user85356 Jul 19 '13 at 16:33
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Write the means $$ \begin{align} H &= \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \\ G &= \sqrt[3]{xyz} \\ A &=\left(\frac{x+y+z}{3}\right) \end{align} $$ then the power mean (AM-GM-HM) inequality gives $H \le G \le A$. Also note $$ xy+yz+zx = xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{3G^3}{H} $$

Now given $$ \begin{align} 8^3 &= (x^2-1)(y^2-1)(z^2-1) \\ &= (x-1)(y-1)(z-1)(x+1)(y+1)(z+1) \\ &= (xyz-(xy+yz+zx)+(x+y+z)-1)(xyz+(xy+yz+zx)+(x+y+z)+1) \\ &= \left(G^3\left(1-\frac{3}{H}\right)+3A-1\right) \left(G^3\left(1+\frac{3}{H}\right)+3A+1\right) \end{align} $$ Then we have either $$ 1-\frac{3}{H} < 0 \\ \implies H < 3 \implies 1< \frac{1}{x}+\frac{1}{y}+\frac{1}{z} $$ or else $$ 1-\frac{3}{H} \ge 0 \\ \implies G^3(1-3/H) \ge H^3(1-3/H) = H^3-3H^2 $$ and hence $$ \begin{align} 8^3 &= \left(G^3\left(1-\frac{3}{H}\right)+3A-1\right) \left(G^3\left(1+\frac{3}{H}\right)+3A+1\right) \\ &\ge (H^3-3H^2+3H-1)(H^3+3H^2+3H+1) \\ &= (H-1)^3(H+1)^3 \\ &= (H^2-1)^3 \\ \implies H &\le 3 \\ \implies 1 &\le \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \end{align} $$

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  • $\begingroup$ One typo: the quadratic mean is a square root, nota square. Although this doesn't affect your results, it is worth noting. Also am gm would work well intead of Cauchy-shwarz $\endgroup$ – chubakueno Jul 20 '13 at 5:07
  • $\begingroup$ Another typo on the final lines: $H \le 3$, Again, this doesnt affect your results $\endgroup$ – chubakueno Jul 20 '13 at 5:15
  • $\begingroup$ One last thing that may be critical and went unnoticed: when you change the sign of the inequality you have, you also change it's direction. That does affect and I couldn't find a fast-repair...If I come up with a solution, i will post it here. Meanwhile, that negative sign is doing our lives impossible. $\endgroup$ – chubakueno Jul 20 '13 at 5:50
  • $\begingroup$ I agree. The inequality involving $-3G^6/H^2$ in the middle does not hold. $\endgroup$ – user85356 Jul 20 '13 at 14:38
  • $\begingroup$ Substantially rewritten. Is there still a problem? $\endgroup$ – Zander Jul 22 '13 at 11:55
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Let $x^2-1=\frac{8bc}{a^2}$, $y^2-1=\frac{8ac}{b^2}$ and $z^2-1=\frac{8ab}{c^2}$, where $a$, $b$ and $c$ are positives.

Hence, by Holder we obtain: $$\sum_{cyc}\frac{1}{x}=\sum_{cyc}\frac{1}{\sqrt{1+\frac{8bc}{a^2}}}=\sum_{cyc}\frac{a}{\sqrt{a^2+8bc}}=$$ $$=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{a}{\sqrt{a^2+8bc}}\right)^2\sum\limits_{cyc}a(a^2+8bc)}{\sum\limits_{cyc}a(a^2+8bc)}}\geq\sqrt{\frac{(a+b+c)^3}{\sum\limits_{cyc}a(a^2+8bc)}}=$$ $$=\sqrt{\frac{\sum\limits_{cyc}(a^3+3a^2b+3a^2c+2abc)}{\sum\limits_{cyc}(a^3+8abc)}}=\sqrt{\frac{\sum\limits_{cyc}(a^3+8abc)+3\sum\limits_{cyc}c(a-b)^2}{\sum\limits_{cyc}(a^3+8abc)}}\geq1.$$ Done!

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By symmetry, the minimum value of the expression

$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$

under the constraint $(x^2-1)(y^2-1)(z^2-1)=8^3$ will occur when $x=y=z$. From this, it follows that the minimum occurs when $x=y=z=3$, and the minimum is $1$.

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  • 1
    $\begingroup$ And by symmetry the minimum of $x^2+y^2+z^2$ when $x^4+y^4+z^4=1$ occurs when... $\endgroup$ – N. S. Jul 19 '13 at 16:19
  • $\begingroup$ Simmetry is a good point to start, but it is not a complete proof. $\endgroup$ – chubakueno Jul 19 '13 at 16:46

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