Giải thích các bước giải:
Xét hiệu:
\(\begin{array}{l}\,\,\,\,\dfrac{{{a^3}}}{{{a^2} + a + 1}} - \dfrac{{2a - 1}}{3}\\ = \dfrac{{3{a^3} - \left( {2a - 1} \right)\left( {{a^2} + a + 1} \right)}}{{3\left( {{a^2} + a + 1} \right)}}\\ = \dfrac{{3{a^3} - 2{a^3} - 2{a^2} - 2a + {a^2} + a + 1}}{{3\left( {{a^2} + a + 1} \right)}}\\ = \dfrac{{{a^3} - {a^2} - a + 1}}{{3\left( {{a^2} + a + 1} \right)}}\\ = \dfrac{{{a^2}\left( {a - 1} \right) - \left( {a - 1} \right)}}{{3\left( {{a^2} + a + 1} \right)}}\\ = \dfrac{{\left( {a - 1} \right)\left( {{a^2} - 1} \right)}}{{3\left( {{a^2} + a + 1} \right)}}\\ = \dfrac{{{{\left( {a - 1} \right)}^2}\left( {a + 1} \right)}}{{3\left( {{a^2} + a + 1} \right)}}\end{array}\)
Ta có: \({\left( {a - 1} \right)^2} \ge 0\,\,\forall a\)
\({a^2} + a + 1 = {a^2} + 2.a.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4} = {\left( {a + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > \,0\,\,\,\forall a\)
\(a \ge 0\,\,\left( {gt} \right) \Rightarrow a + 1 > 0\,\,\forall a \ge 0\).
Vậy \(\dfrac{{{{\left( {a - 1} \right)}^2}\left( {a + 1} \right)}}{{3\left( {{a^2} + a + 1} \right)}} \ge 0\,\,\forall a \ge 0\) hay \(\dfrac{{{a^3}}}{{{a^2} + a + 1}} \ge \dfrac{{2a - 1}}{3}\,\,\forall a \ge 0\) (đpcm).
