Lời giải
Áp dụng BĐT Cauchy, ta có:
\(\frac{{ab}}{{\sqrt {ab + 2c} }} = \frac{{ab}}{{\sqrt {ab + \left( {a + b + c} \right)c} }} = \frac{{ab}}{{\sqrt {ab + ac + bc + {c^2}} }}\)
\( = \frac{{ab}}{{\sqrt {\left( {a + c} \right)\left( {b + c} \right)} }} \le \frac{1}{2}\left( {\frac{{ab}}{{a + c}} + \frac{{ab}}{{b + c}}} \right)\)
Tương tự, ta cũng có:
• \(\frac{{bc}}{{\sqrt {bc + 2a} }} \le \frac{1}{2}\left( {\frac{{bc}}{{a + b}} + \frac{{bc}}{{a + c}}} \right)\);
• \(\frac{{ca}}{{\sqrt {ca + 2b} }} \le \frac{1}{2}\left( {\frac{{ca}}{{a + b}} + \frac{{ca}}{{b + c}}} \right)\)
Cộng theo vế 3 BĐT trên ta có:
\(P \le \frac{1}{2}\left( {\frac{{ab + bc}}{{a + c}} + \frac{{bc + ca}}{{a + b}} + \frac{{ab + ca}}{{b + c}}} \right)\)
\( = \frac{1}{2}\left( {\frac{{b\left( {a + c} \right)}}{{a + c}} + \frac{{c\left( {a + b} \right)}}{{a + b}} + \frac{{a\left( {b + c} \right)}}{{b + c}}} \right)\)
\( = \frac{1}{2}\left( {a + b + c} \right) = \frac{1}{2}\,.\,2 = 1\).
Dấu “=” xảy ra khi \(a = b = c = \frac{2}{3}\).
Vậy GTLN của P là 1 khi \(a = b = c = \frac{2}{3}\).