Let$\ P_0 $be our original proposition and$\ P_1 :a<b $ and$\ P_2:a<c $ the statements that are equivalent to$\ P_0 $. Now, if it is known that$\ b<c $, then there is an interval of values of$\ x $ such that$\ b<x<c $. Since both$\ P_1 $ and$\ P_2 $ are necessary and sufficient for the truth of$\ P_0 $, doesn't it mean that for the latter to be true it is only needed that$\ a $ assumes any value respectively$\ <b $ and$\ <c $, and we can't just say that$\ a$ must be$\ <b $? If so, recall $\ b<x<c $. If$\ a $ is in this interval, then$\ P_1 $ is false and $\ P_2 $ is true, leading to the contradictory result that $\ P_0 $ is at the same time true and false. Thus, may we conclude that$\ P_0 $ is false?
I hope I made myself clear.
EDIT: I'm putting here some more details from my answer to Nick and paw88789. what I'm saying is that, known that $\ P_2 $ is a necessary and sufficient condition for $\ P_0 $, one should be able to pick any $\ x<c $ for $\ P_0 $ to be true. But, as it turns out, it's not like that. In fact, if $\ b<a<c $ we have that $\ P_0 $ is both true and false, which is a contradiction, so they're all false. Alternatively, we may note that it must be $\ a<b $ to avoid the "contradictory interval". Thus, $\ P_2 $ loses its sufficiency, which is absurd since it was demonstrated. This is why I thought of concluding that all the propositions are false.
My latest guess is that since both $\ P_1 $ and $\ P_2 $ have been proven to be equivalent to$\ P_0 $, the truth of either of the three statements implies that of the others, and that's exactly because $\ a $ can't lie in the "contradictory interval", so showing that$\ a <c $ gives us that$\ a <b $ as well. That is, we can't deduce the falsity of the three statements.