Check that: $(1-a)(1-abc) \leq (1-ab)(1-ac)$ and $(1+ab)(1+ac) \leq (1+a)(1+abc)$.

By $a, b, c \in [0;1]$ we have $a(bc+1) \geqslant a(b+c)$ and all $(1-a),(1+ab),(1+ac),(1-abc) , (1+a),(1-ab),(1-ac),(1+abc) \ge 0$ So, $(1-ab)(1-ac)-(1-a)(1-abc) = a(bc+1-b-c) \geqslant 0$ and $(1+a)(1+abc)-(1+ab)(1+ac)=a(bc=1-b-c) \geqslant 0$ P.S. it is same with @aqwxderf

Let $a, b, c \in [1,3]$. Prove that$$(1-a)(1+ab)(1+ac)(1-abc) \leq (1+a)(1-ab)(1-ac)(1+abc) $$Let $a, b, c \in [0,\frac 12]$. Prove that$$(1-a)(1+ab)(1+ac)(1+abc)\leq \frac 53\left(\frac{104}{52-\sqrt{1426-261\sqrt{29}}}-1\right)(1+a)(1-ab)(1-ac)(1-abc) $$

a_507_bc wrote: Prove that for $a, b, c \in [0;1]$, $$(1-a)(1+ab)(1+ac)(1-abc) \leq (1+a)(1-ab)(1-ac)(1+abc).$$

Attachments:

Let $s=1+a^2bc$ so the inequality becomes $(s-a(1+bc))(s+a(b+c))\leq (s+a(1+bc))(s-a(b+c))$. Then, $a(1+bc)\geq a(b+c) \Longrightarrow 1+bc\geq b+c \Longrightarrow b(c-1)+1\geq c$. If $c\neq 1, c-1<0$. Divide both sides by $c-1$, we get $b+\frac{1}{c-1}\leq \frac{c}{c-1} \Longrightarrow b\leq \frac{c-1}{c-1}=1$ which is true by the problem statement. Note that when we divide by a negative, the inequality sign changes. If $c=1, 1+b=1+b$ so the inequality remains true as well.