Решение иррациональных неравенств методом последовательных упрощений

Пример. Решите неравенство $\left(𝑥+1\right)\sqrt{{𝑥}^2+1}\ge {𝑥}^2-1$.

$\left(𝑥+1\right)\sqrt{{𝑥}^2+1}-({𝑥}^2-1)\ge 0$,

$\left(𝑥+1\right)\sqrt{{𝑥}^2+1}-(𝑥-1)(𝑥+1)\ge 0$,

$\left(𝑥+1\right)(\sqrt{{𝑥}^2+1}-\left(𝑥-1\right))\ge 0$,

$\left(𝑥+1\right)(\sqrt{{𝑥}^2+1}-𝑥+1)\ge 0$.

Данное неравенство равносильно совокупности двух систем:

1) $\left\{\begin{array}{c}𝑥+1\ge 0,\\ \sqrt{{𝑥}^2+1}-𝑥+1\ge 0;\end{array}\right.$ $\left\{\begin{array}{c}𝑥\ge -1,\\ \sqrt{{𝑥}^2+1}\ge 𝑥-1;\end{array}\right.$ $\left\{\begin{array}{c}𝑥\ge -1,\\ \left[\begin{array}{c}𝑥-1<0,\\ \left\{\begin{array}{c}𝑥-1\ge 0,\\ {𝑥}^2+1\ge {𝑥}^2-2𝑥+1;\end{array}\right.\end{array}\right.\end{array}\right.$ $\left\{\begin{array}{c}𝑥\ge -1,\\ \left[\begin{array}{c}𝑥<1,\\ \left\{\begin{array}{c}𝑥\ge 1,\\ 𝑥\ge 0;\end{array}\right.\end{array}\right.\end{array}\right.$ $\left\{\begin{array}{c}𝑥\ge -1,\\ \left[\begin{array}{c}𝑥<1,\\ 𝑥\ge 1;\end{array}\right.\end{array}\right.$ $𝑥\ge -1$.

2) $\left\{\begin{array}{c}𝑥+1\le 0,\\ \sqrt{{𝑥}^2+1}-𝑥+1\le 0;\end{array}\right.$ $\left\{\begin{array}{c}𝑥\le -1,\\ \sqrt{{𝑥}^2+1}\le 𝑥-1;\end{array}\right.\left\{\begin{array}{c}𝑥\le -1,\\ {𝑥}^2+1\le {𝑥}^2-2𝑥+1;\end{array}\right.$ $\left\{\begin{array}{c}𝑥\ge -1,\\ 𝑥\le 0;\end{array}\right.$ нет решений.

Ответ: $[-1; +\infty )$.