若a，b，c∈R+,a^2+b^2+c^2=1.求证:
(2-bc)*(2-ca)*(2-ab)≥1

若a，b，c∈R+,a^2+b^2+c^2=3.求证:
(2-bc)*(2-ca)*(2-ab)≥1

简证 齐次后为
(2∑a^2-3bc)*(2∑a^2-3ca)*(2∑a^2-3ab)≥(∑a^2)^3 (1)
<====>
7(∑a^2)^3-12∑bc*(∑a^2)^2+18abc∑a*∑a^2-27(abc)^3≥0

<===>
(∑a^2)[7(∑a^2)^2-12∑bc∑a^2+15abc∑a]+3abc(∑a∑a^2-9abc)≥0

∵
7(∑a^2)^2-12∑bc∑a^2+15abc∑a≥0 (2)
∑a∑a^2-9abc≥0 (3)

(3)式显然成立.
设a=min(a,b,c),(2)式分解为
[7(a-b)(a-c)+11(b-c)^2+2a(b+c)+bc](a-b)(a-c)
+[7(b-c)^2+7bc-3a(b+c)](b-c)^2≥0
显然(2)式成立.从而(1)得证.