证明:(1)已知x,y都是正实数,求证:x3+y3≥x2y+xy2,(2)已知a,b,c∈R+,且a+b+c=1,求证:a2+b2+

2025-05-04 22:13:23
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回答1:

证明:(1)∵(x3+y3 )-(x2y+xy2)=x2 (x-y)+y2(y-x)=(x-y)(x2-y2 )
=(x+y)(x-y)2
∵x,y都是正实数,∴(x-y)2≥0,(x+y)>0,∴(x+y)(x-y)2≥0,
∴x3+y3≥x2y+xy2
(2)∵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

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,当且仅当a=b=c 时,等号成立.