Homogenization and AM-GM.

parmenides51 wrote: Let $x, y, z$ be positive real numbers such that $2(xy + yz + zx) = xyz$. Prove that $\frac{1}{(x-2)(y-2)(z-2)} + \frac{8}{(x+2)(y+2)(z+2)} \le \frac{1}{32}$ $$\iff$$Let $a,b,c>0$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ . Prove that $$\frac{1}{(a-1)(b-1)(c-1)}+\frac{8}{(a+1)(b+1)(c+1)}\leq\frac{1}{4}$$

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parmenides51 wrote: Let $x, y, z$ be positive real numbers such that $2(xy + yz + zx) = xyz$. Prove that $\frac{1}{(x-2)(y-2)(z-2)} + \frac{8}{(x+2)(y+2)(z+2)} \le \frac{1}{32}$ The condition may be rewritten as $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} = \frac{1}{2}$. By AM-GM-HM we have $\frac{x+y+z}{3} \ge \sqrt[3]{xyz} \ge \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} = 6$. Therefore $\sum_{cyc} x \ge 18$, $\sum_{cyc} xy \ge 108$ and $xyz \ge 216$. \[(x-2)(y-2)(z-2) = xyz \, - \, 2 \sum_{cyc} xy \,+ \, 4 \sum_{cyc} x \, - \, 8 = 4 \sum_{cyc} x \, - \, 8 \ge 64\]and \[(x+2)(y+2)(z+2) = xyz \, + \, 2 \sum_{cyc} xy \, + \, 4 \sum_{cyc} x \, + \, 8 \ge 512\]Therefore \[\frac{1}{(x-2)(y-2)(z-2)} + \frac{8}{(x+2)(y+2)(z+2)} \le \frac{1}{64} + \frac{8}{512} = \frac{1}{32}\]

Tell me if there is a flaw anywhere: Take everything to RHS Use Titu's lemma (for 3 terms, the 2 term equivalent) with the - in numerator for 8/(x+2)... term and in denominator for 1/(x-2)... term we get To Prove: 32 - (x-2)(y-2)(z-2) + (x+2)(y+2)(z+2) > 0 (The denominator condition for Titu's Lemma to hold) expanding and using the given equality in question, we get 48 + 2xyz in denominator which is obviously > 0