Lời giải

Áp dụng BĐT Bunhiacopxki, ta có:

\({\left( {\sqrt {\frac{{{a^3}}}{{5{a^2} + {{\left( {b + c} \right)}^2}}}} + \sqrt {\frac{{{b^3}}}{{5{b^2} + {{\left( {c + a} \right)}^2}}}} + \sqrt {\frac{{{c^3}}}{{5{c^2} + {{\left( {a + b} \right)}^2}}}} } \right)^2}\)

\( = {\left( {\sqrt a \sqrt {\frac{{{a^2}}}{{5{a^2} + {{\left( {b + c} \right)}^2}}}} + \sqrt b \sqrt {\frac{{{b^2}}}{{5{b^2} + {{\left( {c + a} \right)}^2}}}} + \sqrt c \sqrt {\frac{{{c^2}}}{{5{c^2} + {{\left( {a + b} \right)}^2}}}} } \right)^2}\)

\( \le \left( {a + b + c} \right)\left( {\frac{{{a^2}}}{{5{a^2} + {{\left( {b + c} \right)}^2}}} + \frac{{{b^2}}}{{5{b^2} + {{\left( {c + a} \right)}^2}}} + \frac{{{c^2}}}{{5{c^2} + {{\left( {a + b} \right)}^2}}}} \right)\)

Ta cần chứng minh: \(\frac{{{a^2}}}{{5{a^2} + {{\left( {b + c} \right)}^2}}} + \frac{{{b^2}}}{{5{b^2} + {{\left( {c + a} \right)}^2}}} + \frac{{{c^2}}}{{5{c^2} + {{\left( {a + b} \right)}^2}}} \le \frac{1}{3}\).

Không mất tính tổng quát ta giả sử

\(a + b + c = 1;\;a \ge b \ge c \Rightarrow a \ge \frac{1}{3} \ge c\)

BĐT trở thành

• Xét \(c \ge \frac{1}{8}\), thì ta có:

\(9 - \sum {\frac{{27{a^2}}}{{6{a^2} - 2a + 1}} = } \sum {\left( {12a - 1 - \frac{{27{a^2}}}{{6{a^2} - 2a + 1}}} \right) = } \sum {\frac{{{{\left( {3a - 1} \right)}^2}\left( {8a - 1} \right)}}{{6{a^2} - 2a + 1}} \ge 0} \)

• Xét \(c \le \frac{1}{8}\), thì ta có:

\(6\left( {VT - VP} \right) = \frac{{6{a^2} - 2a + 1}} + \frac{{6{b^2} - 2b + 1}} + \frac{{6{c^2} - 2c + 1}}\)

\( = \frac{{6{a^2} - 2a + 1}} + \frac{{6{b^2} - 2b + 1}} + \frac{{6{c^2}}}{{6{c^2} - 2c + 1}}\)

\( = \frac{{2{{\left( {a - b} \right)}^2}\left( {3c - 2} \right)}}{{\left( {6{a^2} - 2a + 1} \right)\left( {6{b^2} - 2b + 1} \right)}} + c\left( {\frac{{6{c^2} - 2c + 1}} - \frac{1}{{6{a^2} - 2a + 1}} - \frac{1}{{6{b^2} - 2b + 1}}} \right)\)

Ta cần chứng=  minh \(\frac{1}{{6{a^2} - 2a + 1}} - \frac{1}{{6{b^2} - 2b + 1}} \ge \frac{{6{c^2} - 2c + 1}}\)

Do \(c \le \frac{1}{8} \Rightarrow \frac{{6{c^2} - 2c + 1}} \le 1\)

Suy ra cần chứng minh \(\frac{1}{{6{a^2} - 2a + 1}} - \frac{1}{{6{b^2} - 2b + 1}} \ge 1\)

+) Xét \(b \le \frac{1}{3} \Rightarrow \frac{1}{{6{b^2} - 2b + 1}} \ge 1\)

+) Xét \(b \ge \frac{1}{3}\). Áp dụng BĐT Cauchy ta có:

4 ³ 6(a² + b²) − 2(a + b) + 2

Hay [2(a + b) + c](a + b + c) ³ 3(a² + b²)

Do \(b \ge \frac{1}{3}\) Þ 3b ³ a Þ [2(a + b) + c](a + b + c) ³ 2(a + b)²

= 3(a + b)² + 4ab − a² − b² ³ 3(a² + b²) + 3ab − a² ³ 3(a² + b²).

Vậy BĐT được chứng minh.