作业帮设a,b,c属于R+,且a+b=c,求证a^(2/3)+b^(2/3)大于c^(2/3)
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设a,b,c属于R+,且a+b=c,求证a^(2/3)+b^(2/3)大于c^(2/3)
设a,b,c属于R+,且a+b=c,求证a^(2/3)+b^(2/3)大于c^(2/3)
设a,b,c属于R+,且a+b=c,求证a^(2/3)+b^(2/3)大于c^(2/3)
[c^(2/3)]^3=(a+b)^2=a^2+b^2+2ab
(a^(2/3)+b^(2/3))^3 = a^2 + b^2 + 3a^(4/3) b^(2/3) + 3a^(2/3) b^(4/3)
= c^2 - 2ab + 3a^(4/3) b^(2/3) + 3a^(2/3) b^(4/3)
= c^2 + a^(2/3) b^(2/3) (-2a^(1/3) b^(1/3) + 3a^(2/3) + 3b^(2/3))
=...
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(a^(2/3)+b^(2/3))^3 = a^2 + b^2 + 3a^(4/3) b^(2/3) + 3a^(2/3) b^(4/3)
= c^2 - 2ab + 3a^(4/3) b^(2/3) + 3a^(2/3) b^(4/3)
= c^2 + a^(2/3) b^(2/3) (-2a^(1/3) b^(1/3) + 3a^(2/3) + 3b^(2/3))
= c^2 + a^(2/3) b^(2/3) ((a^(1/3) - b^(1/3))^2 + 2a^(2/3) + 2b^(2/3))
a,b,c 属于R+
所以, (a^(1/3) - b^(1/3))^2 >= 0, 2a^(2/3) > 0, 2b^(2/3) > 0
所以 a^(2/3) b^(2/3) ((a^(1/3) - b^(1/3))^2 + 2a^(2/3) + 2b^(2/3)) > 0
所以 (a^(2/3)+b^(2/3))^3 - (c^(2/3))^3 > 0
即 (a^(2/3)+b^(2/3))^3 > (c^(2/3))^3
所以 a^(2/3)+b^(2/3) > c^(2/3).
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证明（a^(2/3)+b^(2/3))^3大于c^2即可；
（a^(2/3)+b^(2/3))^3=a^2+b^2+3*a^(2/3)*b^(4/3)+3*a^(4/3)*b^(2/3);
又a^2+2*a*b+b^2=c^2;
所以此时证明3*a^(2/3)*b^(4/3)+3*a^(4/3)*b^(2/3)>2*a*b即可；
用算术平均数的知识
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证明（a^(2/3)+b^(2/3))^3大于c^2即可；
（a^(2/3)+b^(2/3))^3=a^2+b^2+3*a^(2/3)*b^(4/3)+3*a^(4/3)*b^(2/3);
又a^2+2*a*b+b^2=c^2;
所以此时证明3*a^(2/3)*b^(4/3)+3*a^(4/3)*b^(2/3)>2*a*b即可；
用算术平均数的知识
3*a^(2/3)*b^(4/3)+3*a^(4/3)*b^(2/3)>=6*a*b>2*a*b
本题得证！
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