a) Chứng minh \({\cos ^2}x + {\cos ^2}\left( {\frac{{2\pi }}{3} + x} \right) + {\cos ^2}\left( {\frac{{2\pi }}{3} - x} \right) = \frac{3}{2}\)b)&n

Ta có: \({\cos ^2}x + {\cos ^2}\left( {\frac{{2\pi }}{3} + x} \right) + {\cos ^2}\left( {\frac{{2\pi }}{3} - x} \right)\)

\(\begin{array}{l}
 = \frac{1}{2}\left( {1 + \cos 2x} \right) + \frac{1}{2}\left[ {1 + \cos \left( {2x + \frac{{4\pi }}{3}} \right)} \right] + \frac{1}{2}\left[ {1 + \cos \left( {\frac{{4\pi }}{3} - 2x} \right)} \right]\\
 = \frac{3}{2} + \frac{1}{2}\left[ {\cos 2x + \cos \left( {2x + \frac{{4\pi }}{3}} \right) + \cos \left( {\frac{{4\pi }}{3} - 2x} \right)} \right]\\
 = \frac{3}{2} + \frac{1}{2}\left[ {\cos 2x + 2\cos 2x\cos \frac{{4\pi }}{3}} \right]\\
 = \frac{3}{2} + \frac{1}{2}\left[ {\cos 2x + 2\cos \left( { - \frac{1}{2}} \right)} \right]\\
 = \frac{3}{2}
\end{array}\)

b) Điều kiện: \(x \ge \frac{{6 + \sqrt {21} }}{3}\)

Bất phương trình đã cho tương đương với:

\(\begin{array}{l}
3{x^2} - 12x + 5 \le {x^3} + {x^2} - 2x - 1 + 2\sqrt {\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)x\left( {x - 2} \right)} \\
 \Leftrightarrow {x^3} - 2{x^2} + 10x - 6 + 2\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} .\sqrt {\left( {{x^2} + x + 1} \right)}  \ge 0\\
 \Leftrightarrow \left( {{x^3} + {x^2} + x} \right) - 3\left( {{x^2} - 3x + 2} \right) + 2\sqrt {{x^2} - 3x + 2} .\sqrt {{x^3} + {x^2} + x}  \ge 0\\
 \Leftrightarrow 1 - 3.\frac{{{x^2} - 3x + 2}}{{{x^3} + {x^2} + x}} + 2\sqrt {\frac{{{x^2} - 3x + 2}}{{{x^3} + {x^2} + x}}}  \ge 0\,\,\left( * \right)
\end{array}\)

Đặt \(\sqrt {\frac{{{x^2} - 3x + 2}}{{{x^3} + {x^2} + x}}}  = t\,\,\left( {t \ge 0} \right)\) thì 

\(\begin{array}{l}
\left( * \right) \Leftrightarrow 1 - 3{t^2} + 2t \ge 0 \Leftrightarrow  - \frac{1}{3} \le t \le 1 \Rightarrow t \le 1\\
 \Leftrightarrow {x^2} - 3x + 2 \le {x^3} + {x^2} + x \Leftrightarrow {x^3} + 4x + 2 \ge 0\left( 1 \right)
\end{array}\)

Nhận thấy (1) nghiệm đúng với mọi \(x \ge \frac{{6 + \sqrt {21} }}{3}\)

Vậy \(S = \left[ {\frac{{6 + \sqrt {21} }}{3}; + \infty } \right)\)