I'm doubtful about the some parts of the solution to this question:
Suppose that the real numbers $a, b, c > 1$ satisfy the condition $$ {1\over a^2-1}+{1\over b^2-1}+{1\over c^2-1}=1 $$ Prove that $$ {1\over a+1}+{1\over b+1}+{1\over c+1}\leq1 $$
The solution says that noticing $a\geq b\geq c$ leads to ${ a-2\over a+1 }\ge{ b-2\over b+1}\ge{ c-2\over c+1}$ and $ { a+2\over a-1 }\le{ b+2\over b-1}\le{ c+2\over c-1} $, But I don't understand how.
If $a\geq b\geq c$, it is quite visible that $a-2\geq b-2\geq c-2$ and $a+1\geq b+1\geq c+1$, but this does not essentially lead to ${ a-2\over a+1 }\ge{ b-2\over b+1}\ge{ c-2\over c+1}$.
Also after this, the solution uses Chebyshev's Inequality: $$ 3\left( \sum_{cyc}{a^2-4\over a^2-1}\right) \leq \left(\sum_{cyc} {a-2\over a+1} \right)\left(\sum_{cyc} {a+2\over a-1} \right) $$ And then states that the LHS is $0$ by hypothesis.
But as Chebyshev's Inequality uses ascending ordered real numbers, this should be false and I also don't know how LHS became $0$.
I think the solution in the book is either flawed or has a typographical error.
Any explanation will be accepted thankfully!
Because for $a\geq b$ we have $$\frac{a-2}{a+1}-\frac{b-2}{b+1}=\frac{3(a-b)}{(a+1)(b+1)}\geq0.$$ The rest is similar.
Also, the sum $\sum\limits_{cyc}{a^2-4\over a^2-1}=\sum\limits_{cyc}\left(\frac{a-2}{a+1}\cdot\frac{a+2}{a-1}\right)$ is the smallest and it gives a proof of Chebyshov by Rearrangement.
Another way: $$1-\sum_{cyc}\frac{1}{a+1}=\sum_{cyc}\left(\frac{1}{3}-\frac{1}{a+1}\right)=$$ $$=\sum_{cyc}\left(\frac{1}{3}-\frac{1}{a+1}-\frac{1}{4}\left(\frac{1}{3}-\frac{1}{a^2-1}\right)\right)=\sum_{cyc}\frac{(a-2)^2}{4(a^2-1)}\geq0.$$