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题目已知a,b,c∈R,a+2b+3c=6,则a~2+4b~2+9c~2的最小值为__.解法1由柯西不等式得(a~2+4b~2+9c~2)(1~2+1~2+1~2)≥(a+2b+3c)~2,所以3(a~2+4b~2+9c~2)≥36,所以a~2+4b~2+9c~2≥12,当a/1=(2b)/1=(3c)/1且a+2b+3c=6,即a=2,6=1,c=2/3时取得最小值.解法2 a~2+2~2≥4a(a=2时取等号),4b~2+2~2≥8b(b=1时取等号),
The problem is that a, b, c∈R, a + 2b + 3c = 6, the minimum value of a ~ 2 + 4b ~ 2 + 9c ~ 2 is __. Solution 1 is derived from the Cauchy inequality (a ~ 2 + 3 (a ~ 2 + 4b ~ 2 + 9c ~ 2) ≧ 36, and therefore, So a ~ 2 + 4b ~ 2 + 9c ~ 2≥12, when a / 1 = 2b / 1 = 3c / 1 and a + 2b + 3c = 6, = 2/3, the minimum value is obtained when the solution 2 a ~ 2 + 2 ~ 2 ≧ 4 a (equals sign when a = 2), 4b ~ 2 + 2 ~ 2 ≧ 8 b