Show that $$1+ac+ab+3a\leq b+c+abc+3bc$$ if $1\leq a\leq bc,$ $1\leq b\leq ac,$ $1\leq c\leq ab.$
$1\leq a$, so $0\leq (a-1)$. Similarly for $b$ and $c$, so we have
$$ 0\leq (a-1)(b-1)(c-1) $$
$a \leq bc$ so $0\leq bc-a$ and
$$ 0 \leq 4(bc-a) $$
Adding these two we get
$$ 0 \leq (a-1)(b-1)(c-1) + 4(bc-a) $$
Multiplying out yields the result:
$$ 0 \leq (abc-ab-ac+a-bc+b+c-1) + 4bc - 4a $$ $$ 0 \leq abc-ab-ac-3a+3bc+b+c-1 $$ $$ 1+ab+ac+3a \leq abc+3bc+b+c $$ As required.
To solve this I took out the $bc$ and $a$ terms, as given that $bc-a$ could be 0, I reasoned that the inequality should hold with them gone. I "guessed" the factorization $(a-1)(b-1)(c-1)$, and put in some extra terms to compensate, which happened to be of the form $bc-a$.