parmenides51 wrote: Prove that for all positive real numbers $x, y$ and $z$, the double inequality $$0 < \frac{1}{x + y + z + 1} -\frac{1}{(x + 1)(y + 1)(z + 1)} \le \frac18$$holds. When does equality hold in the right inequality? (Walther Janous) Let $x+y+z=3u$. Thus, by AM-GM $$\frac{1}{x + y + z + 1} -\frac{1}{(x + 1)(y + 1)(z + 1)}\leq\frac{1}{3u+1}-\frac{1}{(u+1)^3}\leq\frac{1}{8}.$$

The left side is obvious on expanding. Let the numerator of the expression that arises by subtracting the right side by the left side be $f(x, y, z)$. Then we need to show that $f(x, y, z) \ge 0$. Note that $f(x, y, z) - f\left(x, \frac{y + z}{2}, \frac{y + z}{2}\right) = \frac{(x + 1)(y - z)^2(7 - x - y - z)}{4}$. If $x + y + z > 7$, then the LHS is clearly smaller than $\frac{1}{8}$. Otherwise, $f(x, y, z) \ge f\left(x, \frac{y + z}{2}, \frac{y + z}{2}\right)$. Repeating this for the other variables suitably and taking the limit as the number of these operations tends to infinity gives that equality happens when all variables are equal. What remains is a standard calculus exercise.