How do I prove that for $ x,y,z > 0$ $\frac{x}{y+z} + \frac{y}{x+z} + \frac{z}{x+y} > 3/2$? I'm solving a graduate entrance examination problem. 
We are required to establish the inequality using the following result:
for $x,y > 0$, $\frac{x}{y} + \frac{y}{x} > 2$ (1), which is easy to prove as it is equivalent to $(x - y)^2 > 0$.
But when it comes to an inequality combining $x, y, z$, I got stuck as I've tried to develop the expression into one single fraction and obtain something irreducible.
Any hints ? My intuition tells me that for $x,y,z >0$, any fraction of the form $\frac{x}{y+z}$ is greater than 1/2. As there are three fractions of this kind with mute variables playing symmetrical roles, we get: $1/2 + 1/2 + 1/2 = 3/2$.
I just don't figure out how to play with the result (1).

Apply your formula three times:
$$\frac{x+y}{x+z}+\frac{x+z}{x+y}>2$$
$$\frac{y+x}{y+z}+\frac{y+z}{y+x}>2$$
$$\frac{z+x}{z+y}+\frac{z+y}{z+x}>2$$
Combining gives:
$$\frac{x+y}{x+z}+\frac{x+z}{x+y}+\frac{y+x}{y+z}+\frac{y+z}{y+x}+\frac{z+x}{z+y}+\frac{z+y}{z+x}>6$$
$$\frac{x+y+2z}{x+y}+\frac{x+2y+z}{x+z}+\frac{2x+y+z}{y+z}>6$$
$$1+\frac{2z}{x+y}+1+\frac{2y}{x+z}+1+\frac{2x}{y+z}>6$$
$$2\left(\frac{z}{x+y}+\frac{y}{x+z}+\frac{x}{y+z}\right)>3$$
$$\frac{z}{x+y}+\frac{y}{x+z}+\frac{x}{y+z}>\frac{3}{2}$$
