<s>[INST] How to solve this inequality, with the hypothesis more complicated than the conclusion? Given $x,y,z \in \mathbb{R}$ and $x,y,z>2,$ I want to show that if,
$$\frac{1}{x^2-4}+\frac{1}{y^2-4}+\frac{1}{z^2-4} = \frac{1}{7}$$
then, 
$$\frac{1}{x+2} + \frac{1}{y+2} + \frac{1}{z+2} \leq \frac{3}{7}.$$
I follow the solution here: http://artofproblemsolving.com/community/c6h514107_inequality_by_poru_loh
but I don't know how to alter it to fit this problem?!
 [/INST] Here is an adaptation of pi37 answer in your given link.
Note
$$\sum_{cyc}\dfrac{x^2+25}{x^2-4}=\sum_{cyc}\dfrac{x^2-4+29}{x^2-4}=3+\dfrac{29}{7}=\dfrac{50}{7}$$
and use AM-GM $x^2+25\ge 10x$.so 
$$\sum_{cyc}\dfrac{x^2+25}{x^2-4}\ge\dfrac{10x}{x^2-4}\Longrightarrow\sum_{cyc}\dfrac{x}{x^2-4}\le \dfrac{5}{7}$$
and note
$$\sum_{cyc}\dfrac{x}{x^2-4}-\sum_{cyc}\dfrac{2}{x^2-4}=\sum_{cyc}\dfrac{x-2}{x^2-4}=\sum_{cyc}\dfrac{1}{x+2}$$
so
$$\sum_{cyc}\dfrac{1}{x+2}\le\dfrac{5}{7}-\dfrac{2}{7}=\dfrac{3}{7}$$
</s>