求证:a²+b²+c²+d²≥ab+bc+cd+da
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求证:a²+b²+c²+d²≥ab+bc+cd+da
求证:a²+b²+c²+d²≥ab+bc+cd+da
求证:a²+b²+c²+d²≥ab+bc+cd+da
两边都乘以二,然后把右边的都移过来就OK了
就是
2a^2+2b^2+2c^2+2d^2-2ab+2bc+2cd+2da≥0
(a-b)^2+(b-c)^2+(c-d)^2+(a-d)^2≥0
又(a-b)^2≥0
(b-c)^2≥0
(c-d)^2≥0
(a-d)^2≥0
得证
两边乘以2 右边移过来
2a^2+2b^2+2c^2+2d^2-2ab-2bc-2cd-2da≥0
(a-b)^2+(b-c)^2+(c-d)^2+(a-d)^2≥0
(a-b)^2≥0
(b-c)^2≥0
(c-d)^2≥0
(a-d)^2≥0
所以命题成立