✔ 最佳答案
The statement is clearly true for n=1.
Assume now we have k positive numbers, namely, x_1, x_2, ..., x_k,
such that if x_1*x_2*...*x_k=1, then x_1+x_2+...+x_k>=k.
When n=k+1, we now have k+1 positive numbers, namely,
x_1, x_2, ..., x_k, x_(k+1), and we want to show that
if x_1*x_2*...*x_k*x_(k+1)=1, then x_1+x_2+...+x_k+x_(k+1)>=k+1
If all x_i=1, then it is done.
If not all x_i=1, then WLOG, assume x_1<1<x_(k+1)
then since we have [x_1*x_(k+1)]*x_2*...*x_k=1,
i.e. there are k positive numbers, whose product is 1.
Then by induction assumption, the sum of these numbers is >=k,
i.e. x_1*x_(k+1)+x_2+...+x_k>=k
But now since we have x_1<1<x_(k+1), thus by (a),
x_1+x_2+...+x_k+x_(k+1)>=k+1
Therefore the statement is true for n=k+1.
By the principle of M.I., the statement is true for all n>=1.
The "hence" part is leaved for you.
