设a，b，c，为正数且a+b+c=1，求证（a+1⼀a）平方+（b+1⼀b）平方+（c+1⼀c)平方大于等于100⼀3

用柯西不等式
(1+1+1)[(a+1/a)^2+(b+1/b)^2+(c+1/c)^2]>=(a+1/a+b+1/b+c+1/c)^2=(1+1/a+1/b+1/c)^2
(a+b+c)(1/a+1/b+1/c)>=(1+1+1)^2=9
1/a+1/b+1/c>=9
(1+1/a+1/b+1/c)^2>=(1+9)^2=100
(1+1+1)[(a+1/a)^2+(b+1/b)^2+(c+1/c)^2]>=100
(a+a/1)^2+(b+b/1)^2+(c+c/1)^2>=100/3
原式成立

由柯西不等式3[(a+1/a)^2+(b+1/b)^2+(c+/1/c)^2]=(1+1+1)[(a+1/a)^2+(b+1/b)^2+(c+/1/c)^2]>=(a+1/a+b+1/b+c+1/c)^2=[1+(bc+ca+ab)/(abc)]^2
而(bc+ca+ab)/(abc)>=3(bccaab)^(1/3)/(abc)=3/(abc)^(1/3)>=3/[(a+b+c)/3]=9
故(a+1/a)^2+(b+1/b)^2+(c+/1/c)^2>=(1+9)^2/3=100/3
当且仅当a=b=c时，等号成立 a＋b＋c=1
1/a＋1/b＋1/c=（a＋b＋c）/a＋（a＋b＋c）/b＋（a＋b＋c）/c
=1＋b/a＋c/a＋a/b＋1＋c/b＋a/c＋b/c＋1
=（b/a＋a/b）＋（c/a＋a/c）＋（b/c＋c/b）＋3
≥2＋2＋2＋3=9
1/a＋1/b＋1/c≥9.
(a+b+c)(1/a+1/b+1/c)>=(1+1+1)^2=9

由柯西不等式3[(a+1/a)^2+(b+1/b)^2+(c+/1/c)^2]=(1+1+1)[(a+1/a)^2+(b+1/b)^2+(c+/1/c)^2]>=(a+1/a+b+1/b+c+1/c)^2=[1+(bc+ca+ab)/(abc)]^2
而(bc+ca+ab)/(abc)>=3(bccaab)^(1/3)/(abc)=3/(abc)^(1/3)>=3/[(a+b+c)/3]=9
故(a+1/a)^2+(b+1/b)^2+(c+/1/c)^2>=(1+9)^2/3=100/3
当且仅当a=b=c时，等号成立