Question

设a.b.c∈R+且a+b=c,求证a^2/3+b^2/3>c2/3

Answer 1

证明：欲证a^2/3+b^2/3>c^2/3
即证a^2+3a^(4/3)b^(2/3)+3a^(2/3)b^(4/3)+b^2＞c^2
只需证（a+b)^2-2ab+3a^(4/3)b^(2/3)
+3a^(2/3)b^(4/3)+b^2＞c^2
∵a+b=c,∴(a+b)^2=c^2
只需证
3a^(2/3)b^(2/3)[a^(2/3)+b^(2/3)]＞2ab
只需证a^(2/3)+b^(2/3)＞2/3*a^(1/3)b^(1/3)
∵a^(2/3)+b^(2/3)≥2a^(1/3)b^(1/3),
∴a^(2/3)+b^(2/3)＞2/3*a^(1/3)b^(1/3)成立,原不等式得证.

Answer 2

a+b=c
a=c-b
a^2=(c-b)^2
要证明 a^2/3+b^2/3>c2/3 对两边同时3次方
（a^2/3+b^2/3）^3>c2证明（a^2/3+b^2/3）^3>c2就可以了
（a^2/3+b^2/3）^3=a^2+3a^4/3b+3(a^2/3+b^2/3)^2+b^2=a^2+b^2+2ab+3a^4/3b+3(a^2/3+b^2/3)^2-2ab=c^2+3a^4/3b+3(a^2/3+b^2/3)^2-2ab

3a^4/3b+3(a^2/3+b^2/3)^2-2ab>0所以（a^2/3+b^2/3）^3>c2即 a^2/3+b^2/3>c2/3