Cho các số thực dương a, b, c. Chứng minh rằng:\(\...

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

Ta có:

Áp dụng bất đẳng thức Cauchy – Schaw thì:

\(\begin{array}{l}\dfrac{{{a^2}}}{{{a^2} + ab + {b^2}}} + \dfrac{{{c^2}}}{{c\left( {a + b + c} \right)}} \ge \dfrac{{{{\left( {a + c} \right)}^2}}}{{{a^2} + {b^2} + {c^2} + ab + ac + bc}}\\\dfrac{{{b^2}}}{{b\left( {a + b + c} \right)}} + \dfrac{{{c^2}}}{{{c^2} + ac + {a^2}}} \ge \dfrac{{{{\left( {b + c} \right)}^2}}}{{{a^2} + {b^2} + {c^2} + ab + ac + bc}}\\\dfrac{{{a^2}}}{{a\left( {a + b + c} \right)}} + \dfrac{{{b^2}}}{{{b^2} + bc + {c^2}}} \ge \dfrac{{{{\left( {a + b} \right)}^2}}}{{{a^2} + {b^2} + {c^2} + ab + ac + bc}}\\ \Rightarrow \dfrac{{{a^2}}}{{{a^2} + ab + {b^2}}} + \dfrac{{{b^2}}}{{{b^2} + bc + {c^2}}} + \dfrac{{{c^2}}}{{{c^2} + ac + {a^2}}} + \dfrac{{{a^2}}}{{a\left( {a + b + c} \right)}} + \dfrac{{{b^2}}}{{b\left( {a + b + c} \right)}} + \dfrac{{{c^2}}}{{c(a + b + c)}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ge \dfrac{{{{\left( {a + c} \right)}^2} + {{\left( {b + c} \right)}^2} + {{\left( {a + b} \right)}^2}}}{{{a^2} + {b^2} + {c^2} + ab + ac + bc}}\\ \Leftrightarrow \dfrac{{{a^2}}}{{{a^2} + ab + {b^2}}} + \dfrac{{{b^2}}}{{{b^2} + bc + {c^2}}} + \dfrac{{{c^2}}}{{{c^2} + ac + {a^2}}} + \dfrac{{a + b + c}}{{a + b + c}} \ge \dfrac{{{{\left( {a + c} \right)}^2} + {{\left( {b + c} \right)}^2} + {{\left( {a + b} \right)}^2}}}{{{a^2} + {b^2} + {c^2} + ab + ac + bc}}\\ \Leftrightarrow \dfrac{{{a^2}}}{{{a^2} + ab + {b^2}}} + \dfrac{{{c^2}}}{{{c^2} + ac + {a^2}}} + \dfrac{{{b^2}}}{{{b^2} + bc + {c^2}}} + 1 \ge \dfrac{{2\left( {{a^2} + {b^2} + {c^2} + ab + ac + bc} \right)}}{{{a^2} + {b^2} + {c^2} + ab + ac + bc}}\\ \Leftrightarrow \dfrac{{{a^2}}}{{{a^2} + ab + {b^2}}} + \dfrac{{{c^2}}}{{{c^2} + ac + {a^2}}} + \dfrac{{{b^2}}}{{{b^2} + bc + {c^2}}} + 1 \ge 2\\ \Leftrightarrow \dfrac{{{a^2}}}{{{a^2} + ab + {b^2}}} + \dfrac{{{c^2}}}{{{c^2} + ac + {a^2}}} + \dfrac{{{b^2}}}{{{b^2} + bc + {c^2}}} \ge 1\,\,\left( {dpcm} \right)\end{array}\)

Dấu “=” xảy ra khi và chỉ khi \(a = b = c\).