Monotonicity of a function in two variables Consider the function $f(x,y) = \dfrac{ax+by}{x+y}$ , where $x,y>0$.
I'd like to find out over what region $f(x,y)$ is monotonically decreasing 
(i.e. for $x'>x$ and $y'>y$, then $f(x',y')<f(x,y)~)$. Computing the individual partial derivatives, we get - 
$\begin{eqnarray}
\dfrac{\partial f}{\partial x} &=& \dfrac{(a-b)y}{(x+y)^{2}} \\
\dfrac{\partial f}{\partial y} &=& \dfrac{(b-a)x}{(x+y)^{2}}
\end{eqnarray}$
Hence, if $a>b$, $f$ is monotonically increasing over $x$ and decreasing over $y$ (and vice-versa). I'm curious whether the double partial derivative has any significance?
$\begin{eqnarray}
\dfrac{\partial^{2} f}{\partial x \partial y} &=& \dfrac{(a-b)(x-y)}{(x+y)^{3}}
\end{eqnarray}$
Would it be right to say that if $\dfrac{\partial^{2} f}{\partial x \partial y} < 0$ for some region (such as when $x>y$ and $a<b$), then $f$ is monotonically decreasing over this region?
It would be helpful if someone could point me to a reference paper/book for determining the monotonicity of such functions. 
 A: Note that if $b>a$, then $f(x,y) = a + (b-a)\dfrac y{x+y}$. From this we can see that when $x'>x$, we have $f(x',y)<f(x,y)$. But since $\dfrac y{x+y}=1-\dfrac x{x+y}$, we see that $f(x,y')>f(x,y)$. In other words, from algebra we deduce the same thing you deduced "infinitesimally" using partial derivatives.  There is no region with $x,y>0$ on which this function is monotonically decreasing, as you already stated.
You will need both partial derivatives to be negative on your region in order to get monotonicity as you defined it.
A: While @Ted has provided a generalized answer, the following procedure yields a specific condition that can lead to monotonically decreasing $f(x,y)$ as a function of the increments i.e. $\Delta x = (x'-x)$ and $\Delta y = (y'-y)$.
$
\begin{eqnarray}
\Delta f &=& f(x',y') - f(x,y) \\[5pt] &=& \dfrac{a(x+\Delta x)+b(y+\Delta y)}{x+\Delta x + y+\Delta y} -\dfrac{ax+by}{x+ y}  \\[5pt] &=&  \dfrac{a\cdot[(x+\Delta x)(x+y)-x(x+\Delta x + y+\Delta y)] + b\cdot[(y+\Delta y)(x+y)-y(x+\Delta x + y+\Delta y)]}{(x+\Delta x + y+\Delta y)(x+y)} \\[5pt] &=& \dfrac{a \cdot (\Delta x~y - \Delta y~x) + b\cdot ( \Delta y~x - y~\Delta x) }{(x+\Delta x + y+\Delta y)(x+y)} \\[5pt] &=& \dfrac{ (a-b) \cdot (\Delta x~y - \Delta y~x)}{(x+\Delta x + y+\Delta y)(x+y)}
\end{eqnarray}
$
Given that $x,y,\Delta x,\Delta y$ >0, $~~~$then $\Delta f < 0 $ when 
$
\begin{eqnarray}
\text{1.} ~~a<b ~~\text{and}~~ \dfrac{\Delta x}{x} > \dfrac{\Delta y}{y} \\[5pt]
\text{2.} ~~a>b ~~\text{and}~~ \dfrac{\Delta x}{x} < \dfrac{\Delta y}{y}
\end{eqnarray}
$
