Exercise -1.46(New book) (BOMC-2011)

[sm.joty]

Let $x_1,x_2,.......x_n>0$ such that,\[\frac{1}{1+x_1}+\frac{1}{1+x_2}+\frac{1}{1+x_3}+\cdots\cdots+\frac{1}{1+x_n}=1 \]
Prove that , $x_1x_2.........x_n\geq(n-1)^{n}$

[sourav das]

At first tell us about your work and where do you stuck

[sm.joty]

আমার একটা সমাধান আছে । কিন্তু মনে হচ্ছে সমাধানের শেষ কয়েক লাইনে মনে হয় কিছু একটা ভুল আছে। দেখুনতো কি মনে হয়।

Here, $x_1,x_2,........x_n>0$ such that $\frac{1}{1+x_1}+\frac{1}{1+x_2}+\cdots\cdots+\frac{1}{1+x_n}=1$
Let $a_1=1+x_1, a_2=1+x_2,.................a_n=1+x_n$
so we can write,$\frac{1}{a_1}+\frac{1}{a_2}+\cdots\cdots+\frac{1}{a_n}=1$
$\Rightarrow \frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots\cdots+\frac{1}{a_n}}=n$
Now using AM-GM-HM inequality,
$\Rightarrow \frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots\cdots+\frac{1}{a_n}}\leq\sqrt[n]{a_1a_2a_3......a_n}\leq\frac{a_1+a_2+a_3+\cdots\cdots+a_n}{n}$
that means
\[a_1+a_2+\cdots\cdots+a_n\geq n^2\]
\[\Rightarrow (1+x_1)+(1+x_2)+\cdots\cdots+(1+x_n) \geq n^2\]
\[\Rightarrow n+(x_1+x_2+\cdots\cdots+x_n) \geq n^2\]
\[\Rightarrow x_1+x_2+\cdots\cdots+x_n \geq n(n-1)\]
\[\Rightarrow \frac{x_1+x_2+\cdots\cdots+x_n}{n} \geq n-1\]
$\Rightarrow AM \geq n-1$ where AM is the arithmetic mean. So the geometric mean is GM
\[GM=\sqrt[n]{x_1x_2.......x_n}\] And $AM \geq GM$
So, \[\frac{AM}{AM}\geq \frac{n-1}{GM}\]
$\Rightarrow GM\geq n-1$
So,\[x_1x_2x_3.............x_n \geq (n-1)^{n}\]

[sourav das]

$a \geq b$ then $\frac {1}{a} \leq \frac {1}{b}$
Then tell me where is the bug in your solution?

[*Mahi*]

$\Rightarrow AM \geq n-1$... And $AM \geq GM$
So, \[\frac{AM}{AM}\geq \frac{n-1}{GM}\]
$\Rightarrow GM\geq n-1$

You can't divide inequalities.