设a,b,c属于R+且a+b=c,求证a^2/3+b^2/3>c^2/3

a^(2/3)+b^(2/3)>c^(2/3)
<=a^(2/3)+b^(2/3)>(a+b)^(2/3)
<=[a^(2/3)+b^(2/3)]^3>[(a+b)^(2/3)]^3
<=a^2+3*a^(4/3)*b(2/3)+3*a^(2/3)*b^(4/3)+b^2>a^2+2*a*b+b^2
<=3*a^(4/3)*b(2/3)+3*a^(2/3)*b^(4/3)>2*a*b
<=3*a^(2/3)*b^(2/3)*[a^(2/3)+b^(2/3)]>2*a*b
<=3*[a^(2/3)+b^(2/3)]>2*a^(1/3)*b^(1/3)
<=2*[a^(2/3)+b^(2/3)]>-a^(2/3)-b^(2/3)]+2*a^(1/3)*b^(1/3)
<=2*[a^(2/3)+b^(2/3)]>-[a^(1/3)-b^(1/3)]^2
上式，左边大于0，右边小于等于0，显然成立