Question

设a,b,c属于正数。且a+b=c， 求证:a^(2/3)+b^(2/3)>c^(2/3)?

要过程！！ 谢谢

Answer 1

a+b=c
a^2+b^2+2ab=c^2

[a^(2/3)+b^(2/3)]^3
=a^2+3a^(4/3)b^(2/3)+3a^(2/3)b^(4/3)+b^2
[c^(2/3)]^3=c^2=a^2+b^2+2ab

两个相减
=a^2+3a^(4/3)b^(2/3)+3a^(2/3)b^(4/3)+b^2-(a^2+b^2+2ab)
=3a^(4/3)b^(2/3)+3a^(2/3)b^(4/3)-2ab
由均值不等式
3a^(4/3)b^(2/3)+3a^(2/3)b^(4/3)>=2√[3a^(4/3)b^(2/3)*3a^(2/3)b^(4/3)]=2√(9*a^2b^2)=6ab
所以3a^(4/3)b^(2/3)+3a^(2/3)b^(4/3)-2ab>=6ab-2ab=4ab>0

所以[a^(2/3)+b^(2/3)]^3>[c^(2/3)]^3
a^(2/3)+b^(2/3)>c^(2/3)