来自宋志伟的问题
求证√a^2+b^2+c^2/3≥a+b+c/3
求证√a^2+b^2+c^2/3≥a+b+c/3
1回答
2020-04-16 06:17
求证√a^2+b^2+c^2/3≥a+b+c/3
求证√a^2+b^2+c^2/3≥a+b+c/3
左²-右²=(a²+b²+c²)/3-[(1/3)(a+b+c)]²
=(1/9)[3(a²+b²+c²)-(a+b+c)²]
=(1/9)[2a²+2b²+2c²-2ab-2bc-2ac]
=(1/9)[(a-b)²+(b-c)²+(c-a)²]≥0
则:√[(a²+b²+c²)/3]≥(a+b+c)/3