已知a,b,c∈R+,且a+b+c=1.证明:(Ⅰ)a2+b2+c2≥13;(Ⅱ)a2b+b2c+c2a≥1

2025-03-07 08:44:25
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回答1:

解答:证明(Ⅰ)∵a,b,c∈R+,且a+b+c=1,∴1=(a+b+c)2=a2+b2+c2+2ab+2bc+2ac≤3(a2+b2+c2),
∴a2+b2+c2
1
3
,当且仅当a=b=c时,等号成立.    
(Ⅱ)∵
a2
b
+b≥2a,
b2
c
+c≥2b,
c2
a
+a≥2c,
a2
b
+
b2
c
+
c2
a
+a+b+c≥2(a+b+c),
a2
b
+
b2
c
+
c2
a
≥a+b+c=1,
a2
b
+
b2
c
+
c2
a
≥1