Question

Đã trả lời bởi chuyên gia

Answer 1

Lời giải

Áp dụng BĐT Cô-si ta có:

\(\frac{{{a^2}}}{{a + b}} + \frac{{a + b}}{4} \ge 2\sqrt {\frac{{{a^2}}}{{a + b}}.\frac{{a + b}}{4}} = a\).

Tương tự \(\frac{{{b^2}}}{{b + c}} + \frac{{b + c}}{4} \ge 2\sqrt {\frac{{{b^2}}}{{b + c}}.\frac{{b + c}}{4}} = b\).

Và \(\frac{{{c^2}}}{{c + a}} + \frac{{c + a}}{4} \ge 2\sqrt {\frac{{{c^2}}}{{c + a}}.\frac{{c + a}}{4}} = c\).

Suy ra \(\frac{{{a^2}}}{{a + b}} + \frac{{{b^2}}}{{b + c}} + \frac{{{c^2}}}{{c + a}} \ge a - \frac{{a + b}}{4} + b - \frac{{b + c}}{4} + c - \frac{{c + a}}{4}\)

\( \Rightarrow \frac{{{a^2}}}{{a + b}} + \frac{{{b^2}}}{{b + c}} + \frac{{{c^2}}}{{c + a}} \ge \left( {a + b + c} \right) - \left( {\frac{{a + b}}{4} + \frac{{b + c}}{4} + \frac{{c + a}}{4}} \right)\)

\( \Rightarrow \frac{{{a^2}}}{{a + b}} + \frac{{{b^2}}}{{b + c}} + \frac{{{c^2}}}{{c + a}} \ge \frac{{a + b + c}}{2} = \frac{1}{2}\) (đpcm).

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Answer 2