MT274一道漂亮的不等式题

Posted 2021-01-30 青春的记忆

已知$x_1^2+x_2^2+\\cdots+x_6^2=6,x_1+x_2+\\cdots+x_6=0,$证明:$x_1x_2\\cdots x_6\\le\\dfrac{1}{2}$

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.