Question

数学，不等式

已知a,b,c均>0,a+b+c=1,求证：√3a+1+√3b+1+√3c+1≤3√2

Answer 1

应该是√(3a+1)+√(3b+1)+√(3c+1)≤3√2。
因为a，b，c均>0，a+b+c=1，所以，
[√(3a+1)+√(3b+1)+√(3c+1)]²
=(3a+1)+(3b+1)+(3c+1)+2√[(3a+1)(3b+1)]+2√[(3b+1)(3c+1)]+2√[(3c+1)(3a+1)]
≤(3a+1)+(3b+1)+(3c+1)+[(3a+1)+(3b+1)]+[(3b+1)+(3c+1)]+[(3c+1)+(3a+1)]
=3[(3a+1)+(3b+1)+(3c+1)]=3[3(a+b+c+1)]=3[3(1+1)]=18，
故√(3a+1)+√(3b+1)+√(3c+1)≤3√2。

Answer 2

a，b，c＞0，a+b+c=1，则有3a≤1，3b≤1，3c≤1。
√3a+1≤√2，√3b+1≤√2，√3c+1≤√2
√3a+1+√3b+1+√3c+1≤3√2