由 qeypour 於 星期四 九月 08, 2005 9:08 pm
令a1/b1=r,a2/b2=s,a3/b3=t
則0<r<s<t且a1=b1*r,a2=b2*s,a3=b3*t
sqrt[(a1a2+a2a3+a3a1)/(b1b2+b2b3+b3b1)]
=sqrt[(b1b2*r*s+b2b3*s*t+b3b1*r*t)/(b1b2+b2b3+b3b1)].........(1)
(1)<sqrt[(b1b2*t*t+b2b3*t*t+b3b1*t*t)/(b1b2+b2b3+b3b1)]=t=a3/b3
(1)>sqrt[(b1b2*r*r+b2b3*r*r+b3b1*r*r)/(b1b2+b2b3+b3b1)]=r=a1/b1
所以a1/b1<(1)<a3/b3,得證