Let's manipulate the expression a little bit and see what we come up with: we have
[tex]\dfrac{x+z}{y+z}>\dfrac{x}{y} \iff \dfrac{x+z}{y+z}-\dfrac{x}{y}>0 \iff \dfrac{y(x+z)-x(y+z)}{y(y+z)}>0[/tex]
We can simplify the fraction as
[tex]\dfrac{xy+yz-xy-xz}{y(y+z)}=\dfrac{yz-xz}{y(y+z)}=\dfrac{z(y-x)}{y(y+z)}[/tex]
Since both [tex]y[/tex] and [tex]z[/tex] are positive, their sum will be positive as well. In other words, we can rewrite the fraction as
[tex]\underbrace{z}_{>0}\cdot\underbrace{\dfrac{1}{y}}_{>0}\cdot\underbrace{\dfrac{1}{y+z}}_{>0}\cdot (y-x)[/tex]
So, the sign of this fraction depends on the sign of [tex]y-x[/tex]. If its positive, then the whole fraction is positive (product of 4 positive factors). If it's negative, then the whole fraction is negative (product of 3 positive factors and a negative one).
In other words, we arrived to the desired conclusion:
[tex]\dfrac{x+z}{y+z}>\dfrac{x}{y}\iff y-x>0 \iff y>x[/tex]
