Cho $a, b, c > 0$ Chứng minh bất đẳng thức sau :
$$\dfrac{a^3 + b^3 + c^3}{2abc} + \dfrac{a^2 + b^2}{c^2 + ab} + \dfrac{b^2 + c^2}{a^2 + bc} + \dfrac{c^2 + a^2}{b^2 + ca} \ge \dfrac{9}{2}$$
Chọn HSG lớp 12 Thừa Thiên Huế 2008-2009
Một hướng khác.
Ta có: \[VT = \frac{{{a^2}}}{{2bc}} + \frac{{{b^2}}}{{2ac}} + \frac{{{c^2}}}{{2ab}} + \frac{{{a^2} + {b^2}}}{{{c^2} + ab}} + \frac{{{b^2} + {c^2}}}{{{a^2} + bc}} + \frac{{{c^2} + {a^2}}}{{{b^2} + ac}}\]
\[ = \left( {\frac{{{a^2} + bc}}{{2bc}} + \frac{{{b^2} + {c^2}}}{{{a^2} + bc}}} \right) + \left( {\frac{{{b^2} + ac}}{{2ac}} + \frac{{{c^2} + {a^2}}}{{{b^2} + ac}}} \right) + \left( {\frac{{{c^2} + ab}}{{2ab}} + \frac{{{a^2} + {b^2}}}{{{c^2} + ab}}} \right) - \frac{3}{2}\]
\[ \ge \left( {\frac{{{a^2} + bc}}{{2bc}} + \frac{{2bc}}{{{a^2} + bc}}} \right) + \left( {\frac{{{b^2} + ac}}{{2ac}} + \frac{{2ac}}{{{b^2} + ac}}} \right) + \left( {\frac{{{c^2} + ab}}{{2ab}} + \frac{{2ab}}{{{c^2} + ab}}} \right) - \frac{3}{2}\]
\[ \ge 2 + 2 + 2 - \frac{3}{2} = \frac{9}{2} \Rightarrow Q.E.D\]
XONG