一个高中数学不等式的证明

题目：已知$a_1,a_2,\cdots,a_n$为$n$个正数，且$a_1a_2\cdots a_n=1$,

求证:$$2+a_1)(2+a_2)\cdots (2+a_n)\geqslant 3^n.$$

    证明： 由$a_1,a_2,\cdots,a_n$是$n$个正数及均值不等式有

$(2+a_1)\geqslant 2\sqrt{2a_1}$  

$(2+a_2)\geqslant 2\sqrt{2a_2}$

$\cdots \cdots$ 

$(2+a_n)\geqslant 2\sqrt{2a_n}$

     以上式子左右两边相乘，由$a_1a_2\cdots a_n=1$有
     $$(2+a_1)(2+a_2)\cdots (2+a_n)\geqslant (2\sqrt{2})^n \sqrt {a_1a_2\cdots a_n}=(2\sqrt{2})^n $$
     但是题目要求证$(2+a_1)(2+a_2)\cdots (2+a_n)\geqslant 3^n$，而$3^n\geqslant (2\sqrt{2})^n$,因此均值不等式失效.
     观察所求不等式$(2+a_1)(2+a_2)\cdots (2+a_n)$的特征数$3$,联想不等式$$a+b+c\geqslant\sqrt[3]{abc}.$$
     将$(2+a_1)(2+a_2)\cdots (2+a_n)\geqslant 3^n$中的$2$拆分成$(1+1)$,即$$(1+1+a_1)(1+1+a_2)\cdots (1+1+a_n)\geqslant 3^n$$
     不等式左边用$a+b+c\geqslant\sqrt[3]{abc}$有

                                           $(1+1+a_1)\geqslant 3\sqrt[3]{a_1}$  

                                           $(1+1+a_2)\geqslant 3\sqrt[3]{a_2}$

                                                  $\cdots \cdots$ 

                                           $(1+1+a_n)\geqslant 3\sqrt[3]{a_n}$

     以上式子左右两边相乘，由$a_1a_2\cdots a_n=1$有
     $$(1+1+a_1)(1+1+a_2)\cdots (1+1+a_n)\geqslant 3^n\sqrt[3]{a_1a_2\cdots a_n}=3^n$$
     即$$(2+a_1)(2+a_2)\cdots (2+a_n)\geqslant 3^n.$$
 
证明：题目为一命题，可用数学归纳法证明
    (1)当$n=1$时，显然成立;
    (2)假设当$n=k$时命题成立;
    当n=k+1时,
    对$(2+a_1)(2+a_2)\cdots (2+a_{k+1})$中的$a_1,a_2,\cdots,a_{k+1}$满足
    $a_1,a_2,\cdots,a_{k+1}>0$,$a_1a_2\cdots a_{k}=1$,
    则$a_1,a_2,\cdots,a_{k+1}$有
    $$a_i\geqslant 1,a_j\leqslant 1$$
    其中$i,j=1,2,\cdots,k+1$.
    进而
    $$(a_i-1)(a_j-1)\leqslant 0$$ $$a_i+a_j\geqslant a_ia_j+1$$
从而
  
                                                       $(2+a_1)(2+a_2)\cdots (2+a_{k+1})$

                                                    $=(2+a_1)(2+a_2)\cdots(2+a_i)(2+a_j)\cdots(2+a_{k+1})$

                                                    $=(2+a_1)(2+a_2)\cdots(4+2(a_i+a_2)+a_ia_j)\cdots(2+a_{k+1})$

                                                    $\geqslant3(2+a_1)(2+a_2)\cdots(2+a_ia_j)\cdots(2+a_{k+1}) $

 将$a_ia_j$看作$a_m$,$m=1,2,\cdots,k$,用归纳假设有

                                                        $(2+a_1)(2+a_2)\cdots (2+a_{k+1})$

                                                   $\geqslant3(2+a_1)(2+a_2)\cdots(2+a_m)\cdots(2+a_{k+1})$

                                                   $=3^{k+1}. $

  由(1)(2)有$$(2+a_1)(2+a_2)\cdots (2+a_n)\geqslant 3^n$$成立.