Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
3{x^2} + 3x + 4 - \left( {3x + 4} \right)\sqrt {{x^2} + 3} \ge 0\\
\Leftrightarrow 3{x^2} + 3x + 4 \ge \left( {3x + 4} \right)\sqrt {{x^2} + 3} \\
\Leftrightarrow {\left( {3{x^2} + 3x + 4} \right)^2} \ge {\left( {3x + 4} \right)^2}\left( {{x^2} + 3} \right)\\
\Leftrightarrow \left( {9{x^4} + 9{x^2} + 16 + 18{x^3} + 24x + 24{x^2}} \right) \ge \left( {9{x^2} + 24x + 16} \right)\left( {{x^2} + 3} \right)\\
\Leftrightarrow 9{x^4} + 18{x^3} + 33{x^2} + 24x + 16 \ge 9{x^4} + 24{x^3} + 43{x^2} + 72x + 48\\
\Leftrightarrow 6{x^3} + 10{x^2} + 48x + 32 \le 0\\
\Leftrightarrow x \le - 0,72895...
\end{array}\)
