If $x,y,z$ are positive numbers such that $x+y+z = 1$ show that:
$$ \frac{1}{x+y} + \frac{1}{x+z}+ \frac{1}{y+z} + \frac{9}{2} \geq \frac{3}{1 - (\frac{x-y}{2})^2} + \frac{3}{1 - (\frac{y-z}{2})^2} + \frac{3}{1 - (\frac{x-z}{2})^2}$$

**(Hopefully) Correct Solution**

WLOG we may assume that $x \geq y \geq z$. And for now we assume that $2y \geq x+z$ (the case where $2y < x+z$ is handled later below).

By AM-HM:
$$\frac{1}{y+z} + \frac{1}{y+z} + \frac{1}{x+z} \geq \frac{9}{(y+z)+(y+z)+(x+z)} = \frac{9}{1+y+2z} = \frac{9}{2+z-x}$$
$$\frac{1}{y+z} + \frac{1}{x+z} + \frac{1}{x+z} \geq \frac{9}{1+x+2z} = \frac{9}{2+z-y}$$
Adding the two inequalities:
$$ \frac{1}{y+z} + \frac{1}{x+z} \geq \frac{3}{2+z-x} + \frac{3}{2+z-y} $$
(Call this inequality 1).

**Incorrect Solution**

WLOG we may assume that $x \geq y \geq z$. Suppose $a = x-z, b = x-y$ so $a+b = 2x-y-z = 3x-1$. If $a=b=0$ then $x=y=z=1/3$ and equality happens. Thus at least one of $a,b$ is positive, so $a+b > 0$.
Now, by Cauchy:
$$(\frac{a}{a+b} . \frac{1}{y+z} + \frac{b}{a+b} . \frac{3}{2})(\frac{a(y+z)}{a+b} + \frac{2b}{3(a+b)}) \geq (\frac{a}{a+b} + \frac{b}{a+b})^2 = 1$$
So:
$$\frac{a}{a+b} . \frac{1}{y+z} + \frac{b}{a+b} . \frac{3}{2} \geq \frac{3(a+b)}{3a(y+z) + 2b} = \frac{3}{2+z-x}$$
The last equality is equivalent to:
$$ \frac{1(a+b)}{3a(y+z) + 2b} = \frac{1}{2+z-x} $$
$$ \iff (a+b)(2+z-x) = 3a(y+z) + 2b $$
Because $2+z-x \iff 3-2x-y$ and $y+z = 1-x$
$$ \iff (3x-1)(3-2x-y) = 3(x-z)(1-z) + 2(x-y)$$
Upon expanding and rearranging:
$$ \iff 0 = 3(x-1)(x+y+z - 1) $$
which is true.
On the other hand, using the same definition of $a,b$, we can also show:
$$\frac{b}{a+b} . \frac{1}{y+z} + \frac{a}{a+b} . \frac{3}{2} \geq \frac{3(a+b)}{3b(y+z) + 2a} = \frac{3}{2+y-x}$$
(Using similar identity as above)
Adding the two inequalities, we have:
$$ \frac{1}{y+z} + \frac{3}{2} \geq \frac{3}{2+z-x} + \frac{3}{2+y-x}$$
And similarly:
$$ \frac{1}{x+z} + \frac{3}{2} \geq \frac{3}{2+x-y} + \frac{3}{2+z-y}$$
$$ \frac{1}{x+y} + \frac{3}{2} \geq \frac{3}{2+y-z} + \frac{3}{2+z-x}$$
Now all we need to show is that the sum of LHS is equal to the LHS of the given problem. Indeed it is so because:
$$ \frac{3}{2+y-x} + \frac{3}{2+x-y} = \frac{3(2+x-y) + 3(2+y-x)}{2^2 - (x-y)^2} = \frac{12}{4 - (x-y)^2} = \frac{3}{1 - (\frac{x-y}{2})^2}$$
Edit: what is wrong with this proof?