Question

设x1,x2,……,xn是正数，求证（x1+x2+……+xn)(1/x1 +1/x2 +……+1/xn )≥n^2

柯西不等式

Answer 1

数学归纳法：
n=1 x1*1/x1 =1>=1^2=1
n=2,(x1+x2)(1/x1 +1/x2 )=1+x1/x2+x2/x1+1=2+(x1^2+x2^2)/x1x2
>=2+2x1x2/x1x2=2+2=4=2^2
设n=k时结论成立，即（x1+x2+……+xk)(1/x1 +1/x2 +……+1/xk )≥k^2
当n=k+1时
[x1+x2+……+xk+x(k+1)][1/x1 +1/x2 +……+1/xk+1/x(k+1 )]
=(x1+x2+……+xk)(1/x1 +1/x2 +……+1/xk )+ x(k+1)[1/x1 +1/x2 +……+1/xk+1/x(k+1 )]+[x1+x2+……+xk+x(k+1)]1/x(k+1 )
x(k+1)/x1 +x1/x(k+1)= [x(k+1)^2+x1^2]/x1*x(k+1)>=2
同理x(k+1)/x2 +x2/x(k+1)>= 2.....
x(k+1)/x(k+1) +x(k+1)1/x(k+1)>=2
x(k+1)[1/x1 +1/x2 +……+1/xk+1/x(k+1 )]+[x1+x2+……+xk+x(k+1)]1/x(k+1 )>=2(k+1)
[x1+x2+……+xk+x(k+1)][1/x1 +1/x2 +……+1/xk+1/x(k+1 )]
>=k^2+2(k+1)>(k+1)^2 即当n=k+1时也成立
故命题成立，证毕。