# Is the value of this expression is less than $1$?

$$\frac {y_2x_1-y_1x_2}{x_2+y_2+z(x_1+y_1)}$$

$x_i,y_i,z>0$ and I know that the numerator is positive. For the normalized case where $x_i+y_i = 1$, it is simple enough...but is it possible to see some other way ? I guess one can use ${\rm AM}>{\rm GM}$.

\begin{align*} \frac{y_2x_1 - y_1x_2}{x_2+y_2+z(x_1+y_1)} & < 1\\ \Leftrightarrow\qquad y_2x_1 - y_1x_2& < x_2+y_2+z(x_1+y_1) \\ \Leftrightarrow\qquad y_2(x_1-1)&<y_1(z+x_2+1)+x_2+zx_1 \end{align*} Now chose $$x_1 = 2, z = \epsilon, x_2 = \epsilon, y_1 = \epsilon, y_2 = \delta$$ Then the claim is $$\delta < \epsilon(2\epsilon + 1)+\epsilon+2\epsilon = 2\epsilon^2 +4\epsilon$$ Wich can certainly become false for $\delta$ big enough.