证: 已知a>0,b>0,c>0,a+b+c=1 设X=√(3a+2),Y=√(3b+2),Z=√(3c+2) 则t=X+Y+Z X^2=(3a+2),Y^2=(3b+2),Z^2=(3c+2) X^2+Y^2+Z^2=(3a+2)+(3b+2)+(3c+2)=3*(a+b+c)+6=9 ∵(X-Y)^2≥0,(Y-Z)^2≥0,(X-Z)^2≥0 ∴2XY≤X^2+Y^2,2YZ≤Y^2+Z^2,2XZ≤X^2+Z^2 t=X+Y+Z t^2=X^2+Y^2+Z^2+(2XY)+(2YZ)+(2XZ) ≤X^2+Y^2+Z^2+(X^2+Y^2)+(Y^2+Z^2)+(X^2+Z^2) =3*(X^2+...
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