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?!
 A: Here is an adaptation of hyperbolictangent's answer in your given link.
Let $S:= \frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}$. Note that
\begin{align*}
3-4S &= \frac{x^2-4}{x^2-4} + \frac{y^2-4}{y^2-4} + \frac{z^2-4}{z^2-4} - 4 \left(\frac{x-2}{x^2-4}+\frac{y-2}{y^2-4}+\frac{z-2}{z^2-4}\right)\\
&= \frac{(x-2)^2}{x^2-4}+\frac{(y-2)^2}{y^2-4}+\frac{(z-2)^2}{z^2-4}\\
&= \frac{x-2}{x+2} + \frac{y-2}{y+2} + \frac{z-2}{z+2}.
\end{align*}
Thus,
\begin{align*}
\frac{1}{7} \cdot (3-4S)
&= \left(\frac{1}{x^2-4}+\frac{1}{y^2-4}+\frac{1}{z^2-4}\right)
\left(\frac{x-2}{x+2} + \frac{y-2}{y+2} + \frac{z-2}{z+2}\right)\\
&\ge S^2 & \text{Cauchy-Schwarz}
\end{align*}
So we have
$$0 \ge S^2+\frac{4}{7}S-\frac{3}{7} = \left(S+\frac{2}{7}\right)^2 -\frac{25}{49}$$
$$\frac{5}{7} \ge S+\frac{2}{7}$$
$$\frac{3}{7} \ge S$$
A: Just another way is to note that for $t> 2$,
$$f(t) = \left(\frac17-\frac1{t+2}\right)-\frac9{10}\left(\frac1{21}-\frac1{t^2-4} \right)=\frac{(t-5)^2}{10(t^2-4)} \ge 0$$
and the inequality is equivalent to $f(x)+f(y)+f(z) \ge 0$.  So its true and equality is iff $x=y=z=5$.
A: 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
$$\sum_{cyc}\dfrac{1}{x+2}\le\dfrac{5}{7}-\dfrac{2}{7}=\dfrac{3}{7}$$
