若实数abc满足2^a+2^b=2^a+b.2^a+2^b+2c=2a+b+c 则c最大值是多少
2011-08-21 · 知道合伙人教育行家
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2^a + 2^b = 2^(a+b) ......(1)
2^a + 2^b + 2^c = 2^(a+b+c) ......(2)
∵{ √(2^a) - √(2^b) } ≥ 0
∴2^a + 2^b ≥ 2√{2^(a+b)}
又:2^a + 2^b = 2^(a+b)
∴ 2^(a+b) ≥ 2 * √{2^(a+b)},即2^(a+b) ≥ 2 * 2^[(a+b)/2] = 2^[(a+b+2)/2]
∴(a+b) ≥ (a+b+2)/2,2a+2b ≥ a+b+2,a+b ≥ 2
∴2^(a+b) ≥ 2^2 = 4 .....(3)
根据(2):
2^a+2^b+2^c = 2^(a+b+c) = 2^(a+b) * 2^c
2^(a+b) * 2^c - 2^c = 2^a+2^b
2^c = (2^a+2^b)/{2^(a+b)-1} = 2^(a+b)/{2^(a+b)-1} = 1 / {1 - 1/[2^(a+b)] }
∵2^(a+b) ≥ 4
∴ 0 < 1/[2^(a+b)] ≤ 1/4
∴1 > 1 - 1/[2^(a+b)] ≥ 3/4
∴1 < 1 / {1 - 1/[2^(a+b)] } ≤3 /4
即:1 < 2^c ≤ 3/4
c ≤ log 2 (4/3) = log 2 4 - log 2 3 = 2- log 2 (3)
c最大值是2- log 2 (3)
2^a + 2^b + 2^c = 2^(a+b+c) ......(2)
∵{ √(2^a) - √(2^b) } ≥ 0
∴2^a + 2^b ≥ 2√{2^(a+b)}
又:2^a + 2^b = 2^(a+b)
∴ 2^(a+b) ≥ 2 * √{2^(a+b)},即2^(a+b) ≥ 2 * 2^[(a+b)/2] = 2^[(a+b+2)/2]
∴(a+b) ≥ (a+b+2)/2,2a+2b ≥ a+b+2,a+b ≥ 2
∴2^(a+b) ≥ 2^2 = 4 .....(3)
根据(2):
2^a+2^b+2^c = 2^(a+b+c) = 2^(a+b) * 2^c
2^(a+b) * 2^c - 2^c = 2^a+2^b
2^c = (2^a+2^b)/{2^(a+b)-1} = 2^(a+b)/{2^(a+b)-1} = 1 / {1 - 1/[2^(a+b)] }
∵2^(a+b) ≥ 4
∴ 0 < 1/[2^(a+b)] ≤ 1/4
∴1 > 1 - 1/[2^(a+b)] ≥ 3/4
∴1 < 1 / {1 - 1/[2^(a+b)] } ≤3 /4
即:1 < 2^c ≤ 3/4
c ≤ log 2 (4/3) = log 2 4 - log 2 3 = 2- log 2 (3)
c最大值是2- log 2 (3)
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