# Conditions for some inequality

Suppose I have $f(x)A+g(x)B+h(x)C \ge 0$. Here $A,B,C$ can be positive or negative and $f,g,h$ are nonnegative. I would like to obtain a condition for $f,g,$ and $h$ such that $f'(x)A+g'(x)B+h'(x)C \ge 0$. I will appreciate any substantial comments.

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Unless you tell us more about $f$, $g$, $h$ and the $x$-domain $I$ where all this should hold the condition $f'(x)A+g'(x)B+h'(x)C \ge 0$ cannot be transformed into something simpler. If $I$ is a compact interval then the assumption $f(x)A+g(x)B+h(x)C \ge 0$ is useless, because by adding a suitable constant to $f$ you always can force it to hold without changing the derivative(s). –  Christian Blatter Jul 9 '11 at 15:02
$I$ is a compact interval, and $A, B, C$are given. –  user12847 Jul 9 '11 at 15:39

As the problem is currently stated, $A$, $B$, $C$, and $f$, $g$, $h$ are unfortunately not relevant. Let $$W(x)=Af(x)+Bg(x)+Ch(x).$$
The problem states that $W(x)\ge 0$, presumably for all $x$, and asks for conditions under which $W'(x) \ge 0$ for all $x$.
The condition $W(x) \ge 0$ cannot be of much help. We could ask for $W(x)$ to be non-decreasing, but that is really only a minor restatement of $W'(x)\ge 0$. Apart from that sort of thing, there is no nice condition on a general function that will ensure a non-negative derivative. And despite the apparent complexity of $W(x)$, there are no conditions on it apart from $W(x) \ge 0$.
Thank you. I mean, given $A, B$, and $C$, what are the restrictions for $f,g,$ and $h$ that are sufficient for the second inequallity to hold? –  user12847 Jul 9 '11 at 15:41
@user12847: At the level of generality you seek, even if we put $A=B=C=1$ we can say nothing useful. Even if we put $A=1$, $B=C=0$ we can say nothing useful. I am sure this comes from a real problem, with considerable information about the shapes of the various functions. With (a lot of) further information, there is likely to be an answer. –  André Nicolas Jul 9 '11 at 15:53
Let us say $C=0$. Then $fA+gB \ge 0.$ Suppose $A, B \ge 0.$ Then it will give $B \ge (-f/g) A$. Then $f'A -g'(f/g) A \ge 0$ is a sufficient condition. This level of condition is fine. –  user12847 Jul 9 '11 at 16:05
An additional hint to make you see that @user6312's answer is indeed the crux of the matter: the condition $W(x)\ge0$ is invariant if one multiplies $f(x)$, $g(x)$ and $h(x)$ by $\mathrm{e}^{-ux}$, for every fixed $u$, right? But the condition $W'(x)\ge0$ becomes $W'(x)-uW(x)\ge0$. This becomes a huge problem if $u\gg1$. –  Did Jul 9 '11 at 16:06