Question

If $x_1{x}_2{x}_3....x_n=1(x_1>0,i=1,2,.....n),$ prove that $x_1+x_2+.....x_n\ge n(n\ge 2)$

Answer 1

If $x_1{x}_2$ = 1, then
$x_1+x_2=(\sqrt{x_1}-\sqrt{x_2}{)}^2+2\sqrt{x_1{x}_2}\ge 2\sqrt{x_1{x}_2}=2,$
so that the inequality holds true for n = 2.
Now assume that the sum of any m positive numbers whose product is 1 is greater than or equal to m and let $x_1,x_2,....,x_m,x_{m+1}$ be (m + 1) positive integers such that
$x_1.x_2...x_m.x_{m+1}=1$
We shall prove
$x_1+x_2+...++x_m+x_{m+1}=1$
We shall prove
$x_1+x_2+...+x_m+x_{m+1}\ge m+1.$
If each $x_i$(i = 1,2,...,m + 1) is 1, then $x_1+x_2+...x_m+x_{m+1}=m+,$ so that in this case the inequality holds true.
If $x_i^{\prime }s$ are not all 1, then among them there will be a number greater than 1 and a number less than 1. Let
$x_m$ > 1 and $x_{m+1}$ < 1.
We have $x_1{x}_2...x_{m-1}\left(x_m{x}_{m+1}\right)$ = 1.
This is a product of m numbers and so by our induction hypothesis, we can say that
$x_1+x_2+...x_{m-1}{x}_m{x}_{m+1}\le m$
But then
$x_1+x_2+...x_{m-1}+x_m+x_{m+1}$
$\le m-x_m{x}_{m+1}+x_m+x_{m+1}$ by
$=m+1+(x_m-1)(1-x_{m+1})>m+1$
Since $x_m$-1>0 and 1 - $x_{m+1}>0$ by
so that $(x_m-1)(1-x_{m+1})>0.$
This completes the proof.