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