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

`(x+z)/(y+z)>(x)/(y) \iff (x+z)/(y+z)-(x)/(y)>0 \iff (y(x+z)-x(y+z))/(y(y+z))>0`

We can simplify the fraction as

`(xy+yz-xy-xz)/(y(y+z))=(yz-xz)/(y(y+z))=(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{(1)/(y)}_(>0)\cdot\underbrace{(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:

`(x+z)/(y+z)>(x)/(y)\iff y-x>0 \iff y>x`