若abc三个正数成等差数列,公差不为0,正整数n≥2,求证:a^n+c^n≥2(b^n)

即证(a^n+c^n)/2≥[(a+c)/2]^n=(a+c)^n/2^n即证(a^n+c^n)*2^(n-1)≥(a+c)^n(a+c)^n≤∑[C(n,k)*a^(n-k)*c^k]....(k=0,1,2...,n)(a+c)^n≤∑[C(n,k)*a^k*c^(n-k)]....(k=0,1,2...,n)由于C(n,k)*a^(n-k)*c^k+C(n,k)*a^k*c^(n-k)≤C(n,k)*(a^n+c^n)所以2*(a+c)^n≤∑C(n,k)*(a^n+c^n)=2^n*(a^n+c^n)即(a+c)^n≤(a^n+c^n)*2^(n-1) 收起