<s>[INST] Inequality $\sqrt{1-abc}(3-a-b-c)\geq |(1-a)(1-b)(1-c)|$ Let $a,b,c>0$ and $a^2+b^2+c^2+abc=4$. Prove that 
$$\sqrt{1-abc}(3-a-b-c)\geq |(1-a)(1-b)(1-c)|.$$
Note that equality holds trivially when $a=b=c=1$, but it is not the only case. It also holds more generally when $a=x, b=x, c=2-x^2$, where $0<x^2<2$. 
 [/INST] squaring the given inequality we get
$$- \left( ac+b-2 \right)  \left( bc+a-2 \right)  \left( ab+c-2 \right) \geq 0$$
from $$a^2+b^2+c^2+abc=4$$ we get
$$(b-2)(b+2)=-(a^2+c^2+abc)$$
$$b-2=-\frac{a^2+c^2+abc}{b+2}$$
$$ac+b-2=ac-\frac{a^2+c^2+abc}{abc}$$
$$ac+b-2=-\frac{(a-c)^2}{b+2}<0$$
etc. Thus our product above is non negativ.
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