Bài 104658

Question

Cho $2n$ số dương ${a_1},{a_2},...,{a_n},{b_1},{b_2},...,{b_n}$ Chứng minh rằng
$\sqrt[n]{(a_1+b_1)(a_2+b_2)...(a_n+b_n)}\geq \sqrt[n]{a_1a_2...a_n}+\sqrt[n]{b_1b_2...b_n}$

hoàng anh thọ · Answer 1 · 5/18/2012 9:42:32 AM

Bất đẳng thức cần chứng minh tương đương với
    $\frac{{\sqrt[n]{{{{\rm{a}}_{\rm{1}}}{{\rm{a}}_{\rm{2}}}...{{\rm{a}}_{\rm{n}}}}} + \sqrt[n]{{{{\rm{b}}_{\rm{1}}}{{\rm{b}}_{\rm{2}}}...{{\rm{b}}_{\rm{n}}}}}}}{{\sqrt[{\rm{n}}]{{\left( {{a_1} + {b_1}} \right)\left( {{a_2} + {b_2}} \right)...\left( {{a_n} + {b_n}} \right)}}}} \le 1$
Mà    $\frac{{\sqrt[n]{{{{\rm{a}}_{\rm{1}}}{{\rm{a}}_{\rm{2}}}...{{\rm{a}}_{\rm{n}}}}} + \sqrt[n]{{{{\rm{b}}_{\rm{1}}}{{\rm{b}}_{\rm{2}}}...{{\rm{b}}_{\rm{n}}}}}}}{{\sqrt[{\rm{n}}]{{\left( {{a_1} + {b_1}} \right)\left( {{a_2} + {b_2}} \right)...\left( {{a_n} + {b_n}} \right)}}}} $$ = \sqrt[n]{{\frac{{{{\rm{a}}_{\rm{1}}}{{\rm{a}}_{\rm{2}}}...{{\rm{a}}_{\rm{n}}}}}{{\left( {{a_1} + {b_1}} \right)...\left( {{a_n} + {b_n}} \right)}}}} + \sqrt[n]{{\frac{{{{\rm{b}}_{\rm{1}}}{{\rm{b}}_{\rm{2}}}...{{\rm{b}}_{\rm{n}}}}}{{\left( {{a_1} + {b_1}} \right)...\left( {{a_n} + {b_n}} \right)}}}}$
    $ = \sqrt[n]{{\frac{{{a_1}}}{{{a_1} + {b_1}}}.\frac{{{a_2}}}{{{a_2} + {b_2}}}...\frac{{{a_n}}}{{{a_n} + {b_n}}}}} + \sqrt[n]{{\frac{{{b_1}}}{{{a_1} + {b_1}}}.\frac{{{b_2}}}{{{a_2} + {b_2}}}...\frac{{{b_n}}}{{{a_n} + {b_n}}}}}$
    $ \le \frac{1}{n}\left[ {{{\frac{{{a_1}}}{{{a_1} + {b_1}}}+\frac{{{a_2}}}{{{a_2} + {b_2}}}+...+\frac{{{a_n}}}{{{a_n} + {b_n}}}}}} \right] + \frac{1}{n}\left[ {{{\frac{{{b_1}}}{{{a_1} + {b_1}}}+\frac{{{b_2}}}{{{a_2} + {b_2}}}+...+\frac{{{b_n}}}{{{a_n} + {b_n}}}}}} \right]$
    $ = \frac{1}{n}\left[ {\frac{{{a_1} + {b_1}}}{{{a_1} + {b_1}}} + \frac{{{a_2} + {b_2}}}{{{a_2} + {b_2}}} + ... + \frac{{{a_n} + {b_n}}}{{{a_n} + {b_n}}}} \right] = 1$   (ĐPCM)

Bài 104658

1 Đáp án

Lý thuyết liên quan

Bài 104658

1 Đáp án

Liên quan

Lý thuyết liên quan