# Strategies for deriving properties of an expression

For a given $c^*$, suppose that the following system of non-linear equations in $x$ and $y$,

$f(x,y;c)=0\\ g(x,y;c)=0$

possesses a unique solution $(x^*,y^*)$. The equations are such that I do not have $x$ and $y$ as explicit functions of $c$.

Now I have an expression $h(x,y;c)$. I want to prove that $h(x^*,y^*;c^*)>0$

What are some strategies that I can follow? If I just had a function, I could look for other functions to bound it by but since I have these two constraints, I am not sure how to proceed. I would love to get a variety of suggestions to attack this. Thanks.

An example of the kinds of equations I am running into. In this example, $g(x,y;c)$ is the derivative of $f(x,y;c)$ with respect to $x$. The $h(x,y;c)$ are expressions of $\frac{\delta x}{\delta I}$ and $\frac{\delta y}{\delta I}$ where $I$ is some variable that I have suppressed in this question and I got these expressions assuming that Implicit Function Theorem holds.

$f(x,y;c)=12.8428\, -\frac{0.213828 x^{2.97166} \left(-2392.97 (c-0.057)+31.5023 y^{0.15}-0.450832 y\right)}{y^{2.97166}}+\frac{0.213828 y^{1.70499} \left(4170.75 (c-0.057)-47.9187 y^{0.15}+0.32861 y\right)}{x^{1.70499}}-1403.51 c+16.9824 x^{0.15}+0.833333 x$

$g(x,y;c)=0.833333\, -\frac{0.364576 y^{1.70499} \left(4170.75 (c-0.057)-47.9187 y^{0.15}+0.32861 y\right)}{x^{2.70499}}-\frac{0.635424 x^{1.97166} \left(-2392.97 (c-0.057)+31.5023 y^{0.15}-0.450832 y\right)}{y^{2.97166}}+\frac{2.54737}{x^{0.85}}$

Two examples of $h(x,y;c)$

$h_1(x,y;c)=\frac{0.213828 y^{1.70499} \left(0.32861\, -\frac{7.1878}{y^{0.85}}\right)}{x^{1.70499}}+\frac{0.635424 x^{2.97166} \left(-2392.97 (c-0.057)+31.5023 y^{0.15}-0.450832 y\right)}{y^{3.97166}}+\frac{0.364576 y^{0.704995} \left(4170.75 (c-0.057)-47.9187 y^{0.15}+0.32861 y\right)}{x^{1.70499}}-\frac{0.213828 x^{2.97166} \left(\frac{4.72535}{y^{0.85}}-0.450832\right)}{y^{2.97166}}$ $h_2(x,y;c)=\left(\frac{0.364576 y^{1.70499} \left(0.32861\, -\frac{7.1878}{y^{0.85}}\right)}{x^{2.70499}}+\frac{0.213828 y^{1.70499} \left(0.32861\, -\frac{7.1878}{y^{0.85}}\right)}{x^{1.70499}}+\frac{0.635424 x^{2.97166} \left(-2392.97 (c-0.057)+31.5023 y^{0.15}-0.450832 y\right)}{y^{3.97166}}+\frac{0.621599 y^{0.704995} \left(4170.75 (c-0.057)-47.9187 y^{0.15}+0.32861 y\right)}{x^{2.70499}}-\frac{1.88827 x^{1.97166} \left(-2392.97 (c-0.057)+31.5023 y^{0.15}-0.450832 y\right)}{y^{3.97166}}+\frac{0.364576 y^{0.704995} \left(4170.75 (c-0.057)-47.9187 y^{0.15}+0.32861 y\right)}{x^{1.70499}}-\frac{0.213828 x^{2.97166} \left(\frac{4.72535}{y^{0.85}}-0.450832\right)}{y^{2.97166}}+\frac{0.635424 x^{1.97166} \left(\frac{4.72535}{y^{0.85}}-0.450832\right)}{y^{2.97166}}\right)/\left(\left(\frac{0.986175 y^{1.70499} \left(4170.75 (c-0.057)-47.9187 y^{0.15}+0.32861 y\right)}{x^{3.70499}}-\frac{1.25284 x^{0.971661} \left(-2392.97 (c-0.057)+31.5023 y^{0.15}-0.450832 y\right)}{y^{2.97166}}-\frac{2.16526}{x^{1.85}}\right) \left(\frac{0.213828 y^{1.70499} \left(0.32861\, -\frac{7.1878}{y^{0.85}}\right)}{x^{1.70499}}+\frac{0.635424 x^{2.97166} \left(-2392.97 (c-0.057)+31.5023 y^{0.15}-0.450832 y\right)}{y^{3.97166}}+\frac{0.364576 y^{0.704995} \left(4170.75 (c-0.057)-47.9187 y^{0.15}+0.32861 y\right)}{x^{1.70499}}-\frac{0.213828 x^{2.97166} \left(\frac{4.72535}{y^{0.85}}-0.450832\right)}{y^{2.97166}}\right)\right)$

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The answer to such a question largely depends on the actual expressions of $f$, $g$ and $h$, so maybe could you specify them? – wisefool Dec 15 '12 at 1:14
@wisefool Thanks. I just gave one set of actual expressions. – Amatya Dec 15 '12 at 1:31
@wisefool, be careful what you wish for --- you just might get it. – Gerry Myerson Dec 15 '12 at 1:32
@wisefool I made another edit where I describe the exact relationship between $f,g$ and $h$. The $h$ are basically $\frac{\delta x}{\delta I}$ that I have derived from differentiating $f,g$ with respect to some variable $I$ that I have suppressed. – Amatya Dec 15 '12 at 1:46
Have you tried to plot them? For going down from c=0.1235 to c=0.1225 (or around there) f(x,y,c)=g(x,y,c)=0 seem to pass from $0$ intersections to $2$ intersections; there should be a value with only one intersection, but for all those values of $c$, the area $30\le y\le x\le70$ where these intersections lie is contained in $h_1<0$... – wisefool Dec 15 '12 at 2:30