$(\sqrt[]{x + 3} - \sqrt[]{x - 1})(x - 3 + \sqrt[]{x^2 + 2x - 3} \ge 4$
ĐK: $x \ge 1$
PT <=> $\dfrac{4}{\sqrt[]{x + 3} + \sqrt[]{x - 1}}(x - 3 + \sqrt[]{x^2 + 2x - 3}) \ge 4$
<=> $x - 3 + \sqrt[]{x^2 + 2x - 3} \ge \sqrt[]{x + 3} + \sqrt[]{x - 1}$
<=> $\left\{\begin{matrix} x - 3 + \sqrt[]{x^2 + 2x - 3} \ge 0 \\ 2x^2 - 4x + 6 + 2(x - 3)\sqrt[]{x^2 + 2x - 3} \ge 2x + 2 + 2\sqrt[]{x^2 + 2x - 3}\end{matrix}\right.$
<=> $\left\{\begin{matrix} x - 3 + \sqrt[]{x^2 + 2x - 3} \ge 0 \\ 2x^2 - 6x + 4 + 2(x - 4)\sqrt[]{x^2 + 2x - 3} \ge 0\end{matrix}\right.$ <=> $\left\{\begin{matrix} x - 4 + \sqrt[]{x^2 + 2x - 3} \ge - 1 \\ (x - 4 + \sqrt[]{x^2 + 2x - 3})^2 \ge 13\end{matrix}\right.$
<=> $x - 4 + \sqrt[]{x^2 + 2x - 3} \ge \sqrt[]{13}$
<=> $\sqrt[]{x^2 + 2x - 3} \ge \sqrt[]{13} + 4 - x$
Đây là BPT cơ bản dạng: $\sqrt[]{f(x)} \ge g(x)$
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