Question: Prove that $|\sin(z)|\geq|\sin(x)|$  

**My attempt:**  

We know by definition:

> 
> $$\begin{align}\sin(z)&=\frac{\exp(iz)-\exp(-iz)}{2}=\frac{\exp(-y)\exp(ix)-\exp(y)\exp(-iy)}{2}\\&=\frac{\exp(-y)(\cos(x)+i\sin(x))-\exp(y)(\cos(x)-i\sin(x))}{2}\\&=\frac{(\exp(-y)-\exp(y))(\cos(x)-i\sin(y))}{2}\end{align}$$
> 
> 
> 

Then $|\sin(z)|=\dfrac{\exp(-y)-\exp(y)}{2}$.

Moreover using the same argument we have that, $|\sin(x)|=\dfrac{1}{2}$.

Here i'm stuck. Can someone help me?

Answer: 
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.
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}
$