四元不等式

chudengshuxue

$a, b, c, d \inR_+$，且 $a+b+c+d=4$，求证：$a^2 b c+b^2 c d+c^2 d a+d^2 a b \leq 4$

hbghlyj

设$\left\{p,q,r,s\right\}=\left\{a,b,c,d\right\},p\geq q\geq r\geq s$,由排序不等式,$\ a^2bc+b^2cd+c^2da+d^2ab=a(abc)+b(bcd)+c(cda)+d(dab)
\le p\left(pqr\right)+q\left(pqs\right)+r\left(prs\right)+s\left(qrs\right)=\left(pq+rs\right)\left(pr+qs\right)\le\left(\frac{pq+rs+pr+qs}{2}\right)^2=\frac{1}{4}(\left(p+s\right)\left(q+r\right))^2\le\frac{1}{4}\left(\left(\frac{p+q+r+s}{2}\right)^2\right)^2=4.$
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chudengshuxue

回复 2# hbghlyj


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