Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\frac{{{x^3} + x}}{{{{\left( {{x^2} - x + 1} \right)}^2}}} \le 2\\
\Leftrightarrow {x^3} + x \le 2{\left( {{x^2} - x + 1} \right)^2}\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\rm{do}}\,\,{x^2} - x + 1 > 0,\,\,\forall x \Rightarrow \left( {{x^2} - x + 1} \right) > 0,\,\,\,\forall x} \right)\\
\Leftrightarrow x\left( {{x^2} + 1} \right) \le 2.{\left[ {\left( {{x^2} + 1} \right) - x} \right]^2}\\
\Leftrightarrow x.\left( {{x^2} + 1} \right) \le 2.\left[ {{{\left( {{x^2} + 1} \right)}^2} - 2.\left( {{x^2} + 1} \right).x + {x^2}} \right]\\
\Leftrightarrow 2.{\left( {{x^2} + 1} \right)^2} - 4.\left( {{x^2} + 1} \right).x + 2{x^2} \ge x.\left( {{x^2} + 1} \right)\\
\Leftrightarrow 2{\left( {{x^2} + 1} \right)^2} - 5.\left( {{x^2} + 1} \right).x + 2{x^2} \ge 0\\
\Leftrightarrow \left[ {2.\left( {{x^2} + 1} \right) - x} \right].\left[ {\left( {{x^2} + 1} \right) - 2x} \right] \ge 0\\
\Leftrightarrow \left( {2{x^2} - x + 2} \right).\left( {{x^2} - 2x + 1} \right) \ge 0\\
\Leftrightarrow \left( {2{x^2} - x + 2} \right).{\left( {x - 1} \right)^2} \ge 0\\
{\left( {x - 1} \right)^2} \ge 0,\,\,\,\forall x\\
2{x^2} - x + 2 = 2.\left( {{x^2} - \frac{1}{2}x + 1} \right) = 2.{\left( {x - \frac{1}{4}} \right)^2} + \frac{{15}}{8} > 0,\,\,\,\forall x\\
\Rightarrow \left( {2{x^2} - x + 2} \right){\left( {x - 1} \right)^2} \ge 0,\,\,\,\,\forall x
\end{array}\)
Vậy bất phương trình đã cho có nghiệm với \(\forall x \in R\)
