f(x)=f(x0)+f'(x0)(x-x0)+……+fn(x0)/n!*(x-x0)^n+o(x-x0)^n
=fn(x0)/n!*(x-x0)^n+o((x-x0)^n)
n为偶数,则(x-x0)^n>=0.fn(x0)>0 则x0附近f(x)>=f(x0),为极小值;fn(x0)[(x1+x2)/2]^n;
x0时任意x1≠x2,(x1^n+x2^n)/2>[(x1+x2)/2]^n的证明:
取对数,等价于证明f(x1)=ln(x1^n+x2^n)-ln2-nln[x1+x2]+nln2
f'=nx2[x1^(n-1)-x2^(n-1)]/[x1^n+x2^n]
明显f在(0,x2)递减,在(x2,+∞)递增.故f(x)>f(x2)=0 (x≠x2)