已知$x_1^2+x_2^2+\cdots+x_6^2=6,x_1+x_2+\cdots+x_6=0,$证明:$x_1x_2\cdots x_6\le\dfrac{1}{2}$
解答:显然只需考虑2个非负4个非正(或者2非正4非负)的情况.
不妨设$x_1,x_2\ge0;x_3,x_4,x_5,x_6\le0$,记$a_1=x_1,a_2=x_2,a_k=-x_k (k=3,4,5,6)$则题目变为
已知$a_1^2+a_2^2+a_3^2+a_4^2+a_5^2+a_6^2=6,a_1+a_2=a_3+a_4+a_5+a_6$,求证:$a_1a_2\cdots a_6\le\dfrac{1}{2}$
$\because a_1a_2\cdots a_6\le \left(\dfrac{a_1+a_2}{2}\right)^2\left(\dfrac{a_3+a_4+a_5+a_6}{4}\right)^4$
$=\dfrac{1}{4}\left(\dfrac{a_1+a_2}{2}\right)^4\left(\dfrac{a_3+a_4+a_5+a_6}{4}\right)^2$
$\le\dfrac{1}{4^3}(a_1^2+a_2^2)^2(a_3^2+a_4^2+a_5^2+a_6^2)\le\dfrac{1}{2\cdot4^3}\left(\dfrac{2\sum\limits_{i=1}^{6}a_i^2}{3}\right)^3$
$=\dfrac{1}{2}$
当$x_1,x_2,x_3,x_4,x_5,x_6$中两个取$\pm\sqrt{2}$,四个取$\mp\dfrac{\sqrt{2}}{2}$时取到等号.