（a1^n+b1^n)*(a2^n+b2^n)*(a3^n+b3^n)……(an^n+bn^n)>=(a1a2a3…an+b1b2…bn)^n快

Question

（a1^n+b1^n)*(a2^n+b2^n)*(a3^n+b3^n)……(an^n+bn^n)>=(a1a2a3…an+b1b2…bn)^n
快

Answer 1

问题对ai,bi均为正数时才成立.否则,如果ai,bi不全是正数,取 a1=1,b1=-1,n为奇数,这时不等号左边为0；在右边只需取 a2=a3=...=an=1,b2=-1,b3=b4=...=bn=1,那么右边为2^n,0>=2^n,矛盾.下面的证明基于 ai,bi 均为正数.
两边开n次方只需证：
n次根号下[(a1^n+b1^n)(a2^n+b2^n)...(an^n+bn^n)]>=a1a2...an+b1b2...bn
由于两边均为正数,所以将不等号两边同时除以左边只需证：
(a1a2...an+b1b2...bn)/n次根号下[(a1^n+b1^n)(a2^n+b2^n)...(an^n+bn^n)]<=1.
注意到
a1a2...an/n次根号下[(a1^n+b1^n)(a2^n+b2^n)...(an^n+bn^n)]
=n次根号下(a1^n*a2^n*...*an^n)/n次根号下[(a1^n+b1^n)(a2^n+b2^n)...(an^n+bn^n)]
=n次根号下[a1^n/(a1^n+b1^n) * a2^n/(a2^n+b2^n) * ... * an^n/(an^n+bn^n)] (由n元均值不等式)
<=1/n*[a1^n/(a1^n+b1^n) + a2^n/(a2^n+b2^n) + ... + an^n/(an^n+bn^n)]
即 a1a2...an/n次根号下[(a1^n+b1^n)(a2^n+b2^n)...(an^n+bn^n)]
<=1/n*[a1^n/(a1^n+b1^n) + a2^n/(a2^n+b2^n) + ... + an^n/(an^n+bn^n)]
同理,b1b2...bn/n次根号下[(a1^n+b1^n)(a2^n+b2^n)...(an^n+bn^n)]
<=1/n*[b1^n/(a1^n+b1^n) + b2^n/(a2^n+b2^n) + ... + bn^n/(an^n+bn^n)]
两式相加得到
(a1a2...an+b1b2...bn)/n次根号下[(a1^n+b1^n)(a2^n+b2^n)...(an^n+bn^n)]
<=1/n*[(a1^n+b1^n)/(a1^n+b1^n) + (a2^n+b2^n)/(a2^n+b2^n) + ... + (an^n+bn^n)/(an^n+bn^n)]
=1/n*n
=1
即 (a1a2...an+b1b2...bn)/n次根号下[(a1^n+b1^n)(a2^n+b2^n)...(an^n+bn^n)]<=1,所以
a1a2...an+b1b2...bn<=n次根号下[(a1^n+b1^n)(a2^n+b2^n)...(an^n+bn^n)]
从而 (a1a2...an+b1b2...bn)^n<=(a1^n+b1^n)(a2^n+b2^n)...(an^n+bn^n).