Question

Suppose x,y and z are positive real numbers. Prove that x+z/y+z>xy if and only if x Question has also been attached...PLS HELP!!!

Answer 1

Let's manipulate the expression a little bit and see what we come up with: we have

$\dfrac{x+z}{y+z}>\dfrac{x}{y} \iff \dfrac{x+z}{y+z}-\dfrac{x}{y}>0 \iff \dfrac{y(x+z)-x(y+z)}{y(y+z)}>0$

We can simplify the fraction as

$\dfrac{xy+yz-xy-xz}{y(y+z)}=\dfrac{yz-xz}{y(y+z)}=\dfrac{z(y-x)}{y(y+z)}$

Since both $y$ and $z$ are positive, their sum will be positive as well. In other words, we can rewrite the fraction as

$\underbrace{z}_{>0}\cdot\underbrace{\dfrac{1}{y}}_{>0}\cdot\underbrace{\dfrac{1}{y+z}}_{>0}\cdot (y-x)$

So, the sign of this fraction depends on the sign of $y-x$. 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:

$\dfrac{x+z}{y+z}>\dfrac{x}{y}\iff y-x>0 \iff y>x$