# SolveAlways rejects inequality

I have two functions of x, f and g. f[x] = g[x] + 2, but the exact form of g[x] is unknown. It is known that g[x] is well-formed, positive, continuous, differentiable, etc., but an exact representation is unknown.

What I would like to find are the conditions under which f[x] is positive. This is clearly whenever g[x] > -2, however, when I use SolveAlways:

f[x_] = g[x] + 2
SolveAlways[f[x] >= 0, x]


I get:

SolveAlways::eqf: 2 + g[x] < 0 is not a well-formed equation. >>

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Reduce[g[x] + 2 >= 0] – Rojo Jan 4 '14 at 4:38
SolveAlways is not used when inequalities are involved - see the documentation. Reduce[f[x] == g[x] + 2 && f[x] > 0, Reals] will give you what you want. – ciao Jan 4 '14 at 5:19
@rasher Could you provide an answer instead of a comment so I can mark this question as answered? Thanks! For whatever it is worth, I had read the help documentation but I thought that since != were used in the examples then inequalities would be alright as well. – AGB Jan 4 '14 at 5:44
Surely! One moment... – ciao Jan 4 '14 at 5:50

Reduce[f[x] == g[x] + 2 && f[x] > 0, Reals]