Chứng minh rằng: $\ \frac{{ab}}{{a + b + 2c}} + \frac{{bc}}{{b + c + 2a}} + \frac{{ca}}{{c + a + 2b}} \le 1.$
Giải:
Áp dụng BĐT:
$\ \left( {x + y} \right)\left( {\frac{1}{x} + \frac{1}{y}} \right) \ge 4 \Leftrightarrow \frac{1}{{x + y}} \le \frac{1}{4}\left( {\frac{1}{x} + \frac{1}{y}} \right).$
Ta có: $\ {\frac{{ab}}{{a + b + 2c}} = \frac{{ab}}{{\left( {a + c} \right) + \left( {b + c} \right)}} \le \frac{1}{4}\left( {\frac{{ab}}{{a + c}} + \frac{{ab}}{{b + c}}} \right)}.$
$\ {\frac{{bc}}{{b + c + 2a}} = \frac{{bc}}{{\left( {a + b} \right) + \left( {a + c} \right)}} \le \frac{1}{4}\left( {\frac{{bc}}{{a + b}} + \frac{{bc}}{{a + c}}} \right)}.$
$\ {\frac{{ac}}{{a + c + 2b}} = \frac{{ac}}{{\left( {a + b} \right) + \left( {b + c} \right)}} \le \frac{1}{4}\left( {\frac{{ac}}{{a + b}} + \frac{{ac}}{{b + c}}} \right)}.$
$\ { \Rightarrow VT \le \frac{1}{4}\left( {\frac{{bc + ac}}{{a + b}} + \frac{{ab + ac}}{{b + c}} + \frac{{ab + bc}}{{a + c}}} \right)}.$
$\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{4}\left[ {\frac{{c\left( {a + b} \right)}}{{a + b}} + \frac{{a\left( {b + c} \right)}}{{b + c}} + \frac{{b\left( {a + c} \right)}}{{a + c}}} \right] = \frac{1}{4}\left( {a + b + c} \right) = 4 = VP.$
Vậy BĐT được chứng minh. Dấu "=" xảy ra khi và chỉ khi: $\ a = b = c = \frac{4}{3}.$
