求证a1^2+a2^2+a3^2>=a1a2+a2a3+a3a1
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求证a1^2+a2^2+a3^2>=a1a2+a2a3+a3a1
求证a1^2+a2^2+a3^2>=a1a2+a2a3+a3a1
求证a1^2+a2^2+a3^2>=a1a2+a2a3+a3a1
简单点就是证明:a²+b²+c²≥ab+bc+ca
证明如下:
因为(a-b)²≥0
(b-c)²≥0
(c-a)²≥0
上述三个式子相加,得:
(a-b)²+(b-c)²+(c-a)²≥0
2a²+2b²+2c²-2ab-2bc-2ac≥0
即:a²+b²+c²≥ab+bc+ca
证明: a1^2+a2^2+a3^2=(a1+a2)^2-2a1a2+a3^3=(a1+a2+a3)^2-2(a1+a2)a3-2a1a2 =(a1+a2+a3)^-2(a1a3+a2a3+a1a2) =