设x,y,z∈R+,求证:xyz(x+y+z+√(x^2+y^2+z^2))/(x^2+y^2+z^2)(yz+zx+xy)≤(3+√3)/9还有这个:设正数xi∈(0,1)(i=1,2,3……n),满足x1+x2+x3+……+xn=a,求证x1/(1-x1)+x2/(1-x2)+x3/(1-x3)……xn/(1-xn)≥na/(n-a)唉，我在不等式面

Question

设x,y,z∈R+,求证:xyz(x+y+z+√(x^2+y^2+z^2))/(x^2+y^2+z^2)(yz+zx+xy)≤(3+√3)/9
还有这个:
设正数xi∈(0,1)(i=1,2,3……n),满足x1+x2+x3+……+xn=a,求证x1/(1-x1)+x2/(1-x2)+x3/(1-x3)……xn/(1-xn)≥na/(n-a)
唉，我在不等式面前就是个白痴么，还望各位高手帮帮忙了。

Answer 1

1、xyz(x+y+z+√(x^2+y^2+z^2))/(x^2+y^2+z^2)(yz+zx+xy)=(x+y+z+√(x^2+y^2+z^2))/(x^2+y^2+z^2)(1/x+1/y+1/z)=(x+y+z)/(x^2+y^2+z^2)(1/x+1/y+1/z)+√(x^2+y^2+z^2)/(x^2+y^2+z^2)(1/x+1/y+1/z),由于(x^2+y^2+z^2)(1/x+1/y+1/z)≥1/3*(x+y+z)^2*(1/x+1/y+1/z)≥3(x+y+z),所以(x+y+z)/(x^2+y^2+z^2)(1/x+1/y+1/z)≤1/3=3/9；又由于(x^2+y^2+z^2)(1/x+1/y+1/z)=√(x^2+y^2+z^2)*√(x^2+y^2+z^2)*(1/x+1/y+1/z)≥√3/3*√(x^2+y^2+z^2)*(x+y+z)*(1/x+1/y+1/z)≥3√3*√(x^2+y^2+z^2),所以√(x^2+y^2+z^2)/(x^2+y^2+z^2)(1/x+1/y+1/z)≤√(x^2+y^2+z^2)/3√3*√(x^2+y^2+z^2)=√3/9；这两个式子相加即可.
2、有多种方法,法一：x/(1-x)=-1+1/(1-x),函数f(x)=1/(1-x)在区间(0,1)上是下凹的,所以由琴生不等式得∑f(xi)≥nf(∑xi/n)=n^2/(n-a),于是∑xi/(1-xi)≥-n+n^2/(n-a)=na/(n-a).
法二：如果不会的话先百度权方和不等式,由权方和不等式(m=2的情况)得∑1^2/(1-xi)≥(∑1)^2/∑（1-xi)=n^2/(n-a),于是∑xi/(1-xi)≥-n+n^2/(n-a)=na/(n-a).