<s>[INST] Show that $(1+x)(1+z) \leq (1-(x+z))^-1$, when x,z > 0 and (x+z) < 1 I think my proof is wrong because I don't need the $(x+z) < 1$ in my argumentation.
Can somebody tell me where I made a mistake in my transformations?
$$(1+x)(1+z) \leq (1-(x+z))^{-1}$$ <=>
$$(1-(x+z))(1+x)(1+z) \leq 1$$ <=>
$$(1+x)(1+z) - (x+z)(1+x)(1+z) \leq 1$$ <=>
$$(1+x+z+xz) - (1+x+z+xz)(x+z) \leq 1$$ <=>
$$(1+x+z+xz) - (1+(x+z)+xz)(x+z) \leq 1$$ <=>
$$(1+x+z+xz) - ((1+(x+z)+xz)(x+z)) -1 \leq 0$$ <=>
$$1+(x+z)+xz - ((x+z)(x+z)^2(xz(x+z))) {-1} \leq 0$$ <=>
$$xz - (x+z)^2 - (xz(x+z)) \leq 0$$ <=>
$$xz - x^2 - 2xz - z^2 - (xz(x+z)) \leq 0$$ <=>
$$-xz - x^2 - z^2 - (xz(x+z)) \leq 0$$
This is true because x,z are positive numbers.
 [/INST] By Expanding we get
$$-x^2z-xz^2-x^2-z^2-xz+1\le 1$$
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