1.解 由均值不等式得

x^2/(1-x^2)+y^2/(1-y^2)+z^2/(1-z^2)

=[x^2/(1-x^2)+1]+[y^2/(1-y^2)+1]+[z^2/(1-z^2)+1]-3

=1/(1-x^2)+1/(1-y^2)+1/(1-z^2)-3

=3/[(1-x^2)(1-y^2)(1-z^2)]^(1⁄3)-3

=3/[((1-x^2)+(1-y^2)+(1-z^2))/3]-3

=9/[3-(x^2+y^2+z^2)]-3

=9⁄2-3

=3⁄2.

当x^2=y^2=z^2=1/3时,取得等号.

因此,x^2/(1-x^2)+y^2/(1-y^2)+z^2/(1-z^2)的最小值为3/2.

2.猜测原题为下面的不等式问题:

证明：对于任意实数a1,a2,…，an，有

|a1|+|a2|+…+|an|<=√[n((a1)^2+…+(an)^2)].

证明:由柯西不等式,得

|a1|+|a2|+…+|an|=1|a1|+1|a2|+…+1*|an|

<=√[(1^2+1^2+…+1^2)((a1)^2+…+(an)^2)]

=√[n((a1)^2+…+(an)^2)].