Cho a, b, c là các số thực dương. Chứng minh rằng: \(\frac{{{{\rm{a}}^2}}}{{5{{\rm{a}}^2} + {{({\rm{b}} + {\rm{c)}}}^2}}} + \frac{{{{\rm{b}}^2}}}{{5{{\rm{b}}^2} + {{({\rm{c}} + {\rm{a)}}}^2}}} + \frac{{{{\rm{c}}^2}}}{{5{{\rm{c}}^2} + {{({\rm{a}} + {\rm{b)}}}^2}}} \le \frac{1}{3}\)

Lời giải:

Ta có: \(\frac{{{\rm{9}}{{\rm{a}}^2}}}{{5{{\rm{a}}^2} + {{({\rm{b}} + {\rm{c)}}}^2}}} = \frac{{{\rm{9}}{{\rm{a}}^2}}}{{{\rm{(}}{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}) + 2({{\rm{a}}^2} + {\rm{bc)}}}}\)

\( = \frac{{{{({\rm{a}} + 2{\rm{a)}}}^2}}}{{{\rm{(}}{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}) + 2(2{{\rm{a}}^2} + {\rm{bc)}}}}\) \[ \le \frac{{{{\rm{a}}^2}}}{{{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}}} + \frac{{{\rm{4}}{{\rm{a}}^2}}}{{2(2{{\rm{a}}^2} + {\rm{bc)}}}} = \frac{{{{\rm{a}}^2}}}{{{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}}} + \frac{{{\rm{2}}{{\rm{a}}^2}}}{{2{{\rm{a}}^2} + {\rm{bc}}}}\]

Suy ra: \[\sum {\frac{{{{\rm{a}}^2}}}{{5{{\rm{a}}^2} + {{({\rm{b}} + {\rm{c)}}}^2}}}}  \le \frac{1}{9}\sum {\left[ {\frac{{{{\rm{a}}^2}}}{{{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2}}} + \frac{{{\rm{2}}{{\rm{a}}^2}}}{{2{{\rm{a}}^2} + {\rm{bc}}}}} \right] \le \frac{1}{3}} \]