We have the condition $ab+bc+ca<abc$ it is equivalent to $(a-1)(b-1)(c-1)>0$ which force $a>1,b+c<0 , b<1,c<1$ or other similar perms $(a-1)(b-1)(c-1)+abc>0$ $(a-1)(1-b)(1-c)+abc>0$ if b and c have the same sign or one of them zero we are done. If not let $c = -x \ \text{where} \ 0<x , 0<b<x<1 \text{or} 0<b<1<x$ however there is no such integer on that range so we are done

Yes, in fact we can even prove \[ab+bc+ca<2024abc\]under the same condition...

hmm then the simplified question will be .Prove that $abc\geq0$

StarLex1 wrote: hmm then the simplified question will be .Prove that $abc\geq0$ Yes, this is of course what your argument actually proves.

A question : if ab + bc + ca < abc then its also < 2abc , right ?

AymaanAZ wrote: A question : if ab + bc + ca < abc then its also < 2abc , right ? Well, this is exactly what we are asked to prove in this problem (under some conditions on $a,b,c$...).