<s>[INST] Finding a Mistake for a Particular Form of Inequality My book depicts that the following problem uses ${x^3\over (1+y)(1+z)}+{(1+y)\over 8}+{(1+z)\over 8} \ge {3x\over 4} $.

Let $x, y, z$ be positive real numbers such that $xyz = 1$. Prove that $$ {x^3\over (1+y)(1+z)}+{y^3\over (1+z)(1+x)}+{z^3\over (1+x)(1+y)}\geq{3\over 4} $$

and also points the mistake on using $ {x^3\over (1+y)(1+z)}+{(1+y)}+{(1+z)} \ge {3x} $ as the equality can not hold.
Why does the equality does not hold?
 [/INST] The inequality $$ {x^3\over (1+y)(1+z)}+{(1+y)}+{(1+z)} \ge {3x} $$ is indeed true by AM-GM with the equality for
$$ {x^3\over (1+y)(1+z)}=1+y=1+z,$$ which gives $y=z$ and $x^3=(1+y)^3$ or $x=1+y,$ which with the condition $xyz=1$ gives $$(1+y)y^2=1.$$
But in the original inequality the equality occurs for $x=y=z=1$ and we got that your using of AM-GM does not save an equality occurring.
By the way, this AM-GM $${x^3\over (1+y)(1+z)}+{1+y\over 8}+{1+z\over 8} \ge {3x\over 4} $$ saves it.
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