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For $a>0, x,y,z\in\mathbb{R}$

$$|x|<a,\ |y|<a, \ |z|<a$$

Demonstrate this inequality:

$$\frac{|(x+y+z+a^2xyz)|}{|1+a^2(xy+xz+yz)|}<{1\over a}$$

Nathaniel Bubis
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marinaaaa
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    SOme of your formatting has gotten corrupted. – Amzoti Aug 05 '13 at 15:13
  • The conditions on $(x,y,z,a)$ might be that $|x|\lt1/a$, $|y|\lt1/a$, $|z|\lt1/a$. Then the inequality holds. – Did Aug 06 '13 at 07:48
  • And the solution closely mimicks the solution provided to you on your near-duplicate question. Please read and ponder it. – Did Aug 06 '13 at 08:06

1 Answers1

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The inequality cannot be demonstrated, because there are counter-examples: For $x=1$, $y=0$, $z=0$, and all $a > 1$, we have $$\frac{|(x+y+z+a^2xyz)|}{|1+a^2(xy+xz+yz)|} = 1 > \frac{1}{a} \cdot $$

gammatester
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