1. Cho ba số thực dương a,b,c thoả mãn \[ab + bc + ca = 1\]. Tìm
1. Ta có:
\(P = \frac{{2a}}{{\sqrt {1 + {a^2}} }} + \frac{b}{{\sqrt {1 + {b^2}} }} + \frac{c}{{\sqrt {1 + {c^2}} }}\)
\( = \frac{{2a}}{{\sqrt {{a^2} + ab + bc + ca} }} + \frac{b}{{\sqrt {{a^2} + ab + bc + ca} }} + \frac{c}{{\sqrt {{a^2} + ab + bc + ca} }}\)
\( = \frac{{2a}}{{\sqrt {\left( {a + b} \right)\left( {a + c} \right)} }} + \frac{b}{{\sqrt {\left( {b + c} \right)\left( {b + a} \right)} }} + \frac{c}{{\sqrt {\left( {c + a} \right)\left( {c + b} \right)} }}\)
\( = \sqrt {\frac{{2a}}{{a + b}}.\frac{{2a}}{{a + c}}} + \sqrt {\frac{{2b}}{{b + a}}.\frac{b}{{2\left( {b + c} \right)}}} + \sqrt {\frac{{2c}}{{c + a}}.\frac{c}{{2\left( {c + b} \right)}}} \)
\(AM - GM\)
\( \le \frac{1}{2}\left[ {\frac{{2a}}{{a + b}} + \frac{{2a}}{{a + c}} + \frac{{2b}}{{b + a}} + \frac{b}{{2\left( {b + c} \right)}} + \frac{{2c}}{{c + a}} + \frac{c}{{2\left( {c + b} \right)}}} \right]\)
\( = \frac{1}{2}\left[ {\left( {\frac{{2a}}{{a + b}} + \frac{{2b}}{{a + b}}} \right) + \left( {\frac{{2a}}{{a + c}} + \frac{{2a}}{{a + c}}} \right) + \frac{1}{2}\left( {\frac{b}{{b + c}} + \frac{c}{{b + c}}} \right)} \right]\;\)
\( = \frac{1}{2}\left( {2 + 2 + \frac{1}{2}} \right) = \frac{9}{4}\)
Dấu “=” xảy ra khi \(a = \frac{{7\sqrt {15} }}{{15}};b = c = \frac{{\sqrt {15} }}{{15}}\)
Vậy P đạt GTLN là \(\frac{9}{4}\) khi \(a = \frac{{7\sqrt {15} }}{{15}};b = c = \frac{{\sqrt {15} }}{{15}}\)
2. Ta có \(ab + bc + ca + abc \le 4\)
\( \Leftrightarrow abc + 2\left( {ab + bc + ca} \right) + 4\left( {a + b + c} \right) + 8 \le \left( {ab + 2a + 2b + 4} \right) + \left( {bc + 2b + 2c + 4} \right)\)
\( + \left( {ca + 2c + 2a + 4} \right)\)
\( \Leftrightarrow \left( {a + 2} \right)\left( {b + 2} \right)\left( {c + 2} \right) \le \left( {a + 2} \right)\left( {b + 2} \right) + \left( {b + 2} \right)\left( {c + 2} \right)\)
\( + \left( {c + 2} \right)\left( {a + 2} \right)\)
\( \Leftrightarrow 1 \le \frac{1}{{a + 2}} + \frac{1}{{b + 2}} + \frac{1}{{c + 2}} \Leftrightarrow \frac{2}{{a + 2}} + \frac{2}{{b + 2}} + \frac{2}{{c + 2}}\;\; \ge 2\)
\( \Leftrightarrow \left( {1 - \frac{2}{{a + 2}}} \right) + \left( {1 - \frac{2}{{b + 2}}} \right) + \left( {1 - \frac{2}{{c + 2}}} \right) \le 1\)
\( \Leftrightarrow 1 \ge \frac{a}{{a + 2}} + \frac{b}{{b + 2}} + \frac{c}{{c + 2}}\)
Áp dụng BĐT Cauchy-Schwarz
\(1 \ge \frac{a}{{a + 2}} + \frac{b}{{b + 2}} + \frac{c}{{c + 2}} = \frac{{{a^2}}}{{{a^2} + 2a}} + \frac{{{b^2}}}{{{b^2} + 2b}} + \frac{{{c^2}}}{{{c^2} + 2c}}\)
\(C - S \ge \frac{{{{\left( {a + b + c} \right)}^2}}}{{{a^2} + {b^2} + {c^2} + 2\left( {a + b + c} \right)}}\)
\( \Rightarrow {\left( {a + b + c} \right)^2} \le \left( {{a^2} + {b^2} + {c^2}} \right) + 2\left( {a + b + c} \right)\)
\( \Leftrightarrow a + b + c \ge ab + bc + ca\) (đpcm)