# Resolve a conditional expression including Exists and ForAll

I don't know the Resolve function exactly, especially when involved with Exists and ForAll, could you please help me in the following example?

$\forall C$, can we find some $x>0$, such that $$c(x)=(C+2x)x^2+C^2x+2Cx^2+C\geq0,$$ and $$C<-2x$$ holds at the sametime?

Exists[x,  x > 0 && ForAll[c, c < -2 x && (c + 2 x) x^2 + c^2 x + 2 c x^2 >= 0]]
Resolve[%]

(* False *)


Edit: In thinking about this, I perhaps took the question as stated too literally and that's not what you meant, i.e., it is obvious that it is not true that for all C, C <-2 X, which the above shows. However, if you meant (and I think this true) for all C with the condition C < -2 X, that's done this way:

Exists[x, x > 0, ForAll[c, c < -2 x, (c + 2 x) x^2 + c^2 x + 2 c x^2 >= 0]]
Reduce[%]

(* True *)


You can use FindInstance to get instance(s), like:

FindInstance[x > 0 && c < -2 x && (c + 2 x) x^2 + c^2 x + 2 c x^2 >= 0, {x, c}, Reals]

(*  {{x -> 1, c -> -4}}  *)

• @van abel: thanks for accept, please check update, I might have taken your question too literally... and there was a typo (cx instead of multiplication c x)
– ciao
Mar 10, 2014 at 6:28