- 设a、b、c、d∈R+,a+b+c+d=1,则√(4a+1)+√
- 则√(4a+1)+√(4b+1)+√(4c+1)+√(4d+1)<6.
- 证明:
变形后可用权方和不等式:
√(4a+1)+√(4b+1)+√(4c+1)+√(4c+1)
=[(4a+1)^(1/2)]/1^(-1/2)+[(4b+1)^(1/2)]/1^(-1/2)+[(4c+1)^(1/2)]/1^(-1/2)+[(4c+1)^(1/2)]/1^(-1/2)
≤[(4a+1)+(4b+1)+(4c+1)+(4d+1)]^(1/2)/(1+1+1+1)^(-1/2)
=[4(a+b+c+d)+4]^(1/2)/4^(-1/2)
=4√2
<6,
故命题得证.