- 若a、b、c∈(0,+∞),且kabc/(a+b+c)≤(a+b?
- 求k的最大值.
- 由均值不等式得,
(a+b)^2+(a+b+4c)^2
=(a+b)^2+[(a+2c)+(b+2c)]^2
≥[2根(ab)]^2+[2根(2ac)+2根(2bc)]^2
=4ab+8ac+8bc+16c根(ab)
于是,
{[(a+b)^2+(a+b+4c)^2]/abc}*(a+b+c)
≥[4ab+8ac+8bc+16c根(ab)](a+b+c)/abc]
=[4/c+8/b+8/a+16/根(ab)](a+b+c)
=8[1/(2c)+1/b+1/a+1/根(ab)+1/根(ab)](a/2+a/2+b/2+b/2+c)
≥8[5(1/2a^2b^2c)^(1/5),5(a^2b^2c/16)^(1/5)]
=100
∴当且仅当a=b=2c>0时取等号,
故k的最大值为100.