<s>[INST] Does $a c>b^2$ if and only if $(1-a) (1-c)<(1-b)^2$, where $0Does $$a c>b^2$$  if and only if  $$(1-a) (1-c)<(1-b)^2,$$ where $0<a<1, 0<c<1$, and $0<b<1$ If it is correct, then how to prove it?
 [/INST] I finally find a easy way to prove it.
First, $ac>b^2$ implies $b^2<(\frac{a+c}{2})^2$. And note that $$(1-a)(1-c)-(1-b)^2 \leq (1-\frac{a+c}{2})^2-(1-b)^2=(2-\frac{a+c}{2}-b)(b-\frac{a+c}{2})<0$$
The last inequality is from $b^2\leq(\frac{a+c}{2})^2$ combined with $a,b,c,\in (0,1)$.
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