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$bpt\Leftrightarrow 3x^2+3x+4 \ge (3x+2)\sqrt{x^2+3}$$\Leftrightarrow \begin{cases}(3x^2+3x+4)^2-(3x+2)^2(x^2+3) \ge0 \\ x \ge \dfrac{-2}3 \end{cases}$$\Leftrightarrow \begin{cases}3x^3+x^2-6x+2 \ge0 \\ x \ge -\dfrac 23 \end{cases}$$\Leftrightarrow \begin{cases}(x-1)(3x^2+4x-2) \ge0 \\ x \ge -\dfrac 23\end{cases}$$\Leftrightarrow \begin{cases}\left[ \begin{array}{l} x \ge 1 \\ \dfrac{-\sqrt{10}-2}{3} \le x \le \dfrac{\sqrt{10}- 2}{3} \end{array} \right. \\x \ge \dfrac{-2}3 \end{cases}$$\Leftrightarrow \left[ \begin{array}{l} x \ge 1\\ \dfrac{-2}{3} \le x \le \dfrac{\sqrt{10}-2}{3} \end{array} \right.$

$bpt\Leftrightarrow 3x^2+3x+4 \ge (3x+2)\sqrt{x^2+3}$Với $x < \frac {-2}3$ thì $VT>0$, $VP<0$ ( đúng)Với $x \ge \frac{-2}3$bpt $\Leftrightarrow 3x^2+3x+4 \ge (3x+2)\sqrt{x^2+3}$$\Leftrightarrow (3x^2+3x+4)^2-(3x+2)^2(x^2+3) \ge0$$\Leftrightarrow 3x^3+x^2-6x+2 \ge0$$\Leftrightarrow (x-1)(3x^2+4x+2) \ge0$$\Leftrightarrow \left[ \begin{array}{l} x \ge 1\\ \dfrac{-\sqrt{10}-2}3 \le x \le \dfrac{\sqrt{10}-2}{3} \end{array} \right.$Kết hợp nghiệm đc $x \in \left(- \infty ;\frac{\sqrt{10}-2}3 \right] \cup [1;+\infty)$

$bpt\Leftrightarrow 3x^2+3x+4 \ge (3x+2)\sqrt{x^2+3}$$\Leftrightarrow \begin{cases}(3x^2+3x+4)^2-(3x+2)^2(x^2+3) \ge0 \\ x \ge \dfrac{-2}3 \end{cases}$$\Leftrightarrow \begin{cases}3x^3+x^2-6x+2 \ge0 \\ x \ge -\dfrac 23 \end{cases}$$\Leftrightarrow \begin{cases}(x-1)(3x^2+4x-2) \ge0 \\ x \ge -\dfrac 23\end{cases}$$\Leftrightarrow \begin{cases}\left[ \begin{array}{l} x \ge 1 \\ \dfrac{-\sqrt{10}-2}{3} \le x \le \dfrac{\sqrt{10}- 2}{3} \end{array} \right. \\x \ge \dfrac{-2}3 \end{cases}$$\Leftrightarrow \left[ \begin{array}{l} x \ge 1\\ \dfrac{-2}{3} \le x \le \dfrac{-\sqrt{10}-2}{3} \end{array} \right.$

$bpt\Leftrightarrow 3x^2+3x+4 \ge (3x+2)\sqrt{x^2+3}$$\Leftrightarrow \begin{cases}(3x^2+3x+4)^2-(3x+2)^2(x^2+3) \ge0 \\ x \ge \dfrac{-2}3 \end{cases}$$\Leftrightarrow \begin{cases}3x^3+x^2-6x+2 \ge0 \\ x \ge -\dfrac 23 \end{cases}$$\Leftrightarrow \begin{cases}(x-1)(3x^2+4x-2) \ge0 \\ x \ge -\dfrac 23\end{cases}$$\Leftrightarrow \begin{cases}\left[ \begin{array}{l} x \ge 1 \\ \dfrac{-\sqrt{10}-2}{3} \le x \le \dfrac{\sqrt{10}- 2}{3} \end{array} \right. \\x \ge \dfrac{-2}3 \end{cases}$$\Leftrightarrow \left[ \begin{array}{l} x \ge 1\\ \dfrac{-2}{3} \le x \le \dfrac{\sqrt{10}-2}{3} \end{array} \right.$

$bpt\Leftrightarrow 3x^2+3x+4 \ge (3x+2)\sqrt{x^2+3}$$\Leftrightarrow \begin{cases}(3x^2+3x+4)^2-(3x+2)^2(x^2+3) \ge0 \\ x \ge \frac{-2}3 \end{cases}$$\Leftrightarrow \begin{cases}15x^3+2x^2+30x+2 \ge0 \\ x \ge -\frac 23 \end{cases}$$\Leftrightarrow \begin{cases}(15x+1)(x^2+2) \ge0 \\ x \ge -\frac 23\end{cases}\Leftrightarrow x \ge -\frac 1{15}$

$bpt\Leftrightarrow 3x^2+3x+4 \ge (3x+2)\sqrt{x^2+3}$$\Leftrightarrow \begin{cases}(3x^2+3x+4)^2-(3x+2)^2(x^2+3) \ge0 \\ x \ge \dfrac{-2}3 \end{cases}$$\Leftrightarrow \begin{cases}3x^3+x^2-6x+2 \ge0 \\ x \ge -\dfrac 23 \end{cases}$$\Leftrightarrow \begin{cases}(x-1)(3x^2+4x-2) \ge0 \\ x \ge -\dfrac 23\end{cases}$$\Leftrightarrow \begin{cases}\left[ \begin{array}{l} x \ge 1 \\ \dfrac{-\sqrt{10}-2}{3} \le x \le \dfrac{\sqrt{10}- 2}{3} \end{array} \right. \\x \ge \dfrac{-2}3 \end{cases}$$\Leftrightarrow \left[ \begin{array}{l} x \ge 1\\ \dfrac{-2}{3} \le x \le \dfrac{-\sqrt{10}-2}{3} \end{array} \right.$