设n>=2,x1,x2,……xn为正数，并且x1+x2+……+ xn=1

∑(i=1,..,n)[xi/√(1-xi)]>=(∑(i=1,..,n)√xi)/√(n-1)

由柯西不等式
∑(i=1,..,n)√xi<=√[(∑(i=1,..,n)1)(∑(i=1,..,n)xi)]=√n

即需证∑(i=1,..,n)[xi/√(1-xi)]>=√n/√(n-1)

由柯西不等式
{∑(i=1,..,n)[1/√(1-xi)]}{∑(i=1,..,n)[√(1-xi)]}>=n^2
∑(i=1,..,n)√(1-xi)<=√[(∑(i=1,..,n)1)(∑(i=1,..,n)1-xi)]=√[n(n-1)]

∑(i=1,..,n)[xi/√(1-xi)]
=∑(i=1,..,n)[1/√(1-xi)]-∑(i=1,..,n)√(1-xi)
>=n^2/{∑(i=1,..,n)[√(1-xi)]}-√[n(n-1)]
>=n^2/√[n(n-1)]-√[n(n-1)]
=√n/√(n-1)

所以原不等式得证