Proof for an inequality problem using the rearrangement inequality

Question

I have found this problem in a book and would like someone to give me a good and elegant proof for it.

The problem is:

If $a_1,a_2,...,a_n$ are positive real numbers and $s=a_1+a_2+...+a_n$, then

$$\frac{a_1}{s-a_1}+\frac{a_2}{s-a_2}+...+\frac{a_n}{s-a_n} \ge \frac{n}{n-1}$$

The book does not provide solutions but only gives hints, which is that it can be solved like another problem using n variables

The problem is:

If $a,b,c$ are the lengths of the sides of a triangle, prove that

$$\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \ge 3$$

Again only hints are given and this is the solution I am proposing:

let $(b_1,b_2,b_3)=\frac{a}{b+c-a},\frac{b}{c+a-b},\frac{c}{a+b-c}$

Let $(a_1,a_2,a_3)=a,b,c$

By the rearrangement inequality we have,

$$a_1b_1+a_2b_2+a_3b_3 \ge a_2b_1 +a_3b_2+a_1b_3$$ $$a_1b_1+a_2b_2+a_3b_3 \ge a_3b_1 +a_1b_2+a_2b_3$$

$$\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \ge \frac{b}{b+c-a}+\frac{c}{c+a-b}+\frac{a}{a+b-c}$$ $$\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c} \ge \frac{c}{b+c-a}+\frac{a}{c+a-b}+\frac{b}{a+b-c}$$

Let $$Q = \frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}$$

Now by adding the two inequalities we have,

$$2Q \ge \frac{b}{b}+\frac{b}{c}-\frac{b}{a}+\frac{c}{c}+\frac{c}{a}-\frac{c}{b}+\frac{a}{a}+\frac{a}{b}-\frac{a}{c}+\frac{c}{b}+\frac{c}{c}-\frac{c}{a}+\frac{a}{c}+\frac{a}{a}-\frac{a}{b}+\frac{b}{a}+\frac{b}{b}-\frac{b}{c}=6+\frac{a-b-a+b}{a}+\frac{a-c-a+c}{b}+\frac{b-a-b+a}{c}=6$$

$$Q \ge 3$$

I do understand the general idea but what is needed in the problem above is a generalization of the proof I provided. However I fail at producing one each time I try. Can someone please provide me with one that is simple and easy to read. Thanks.

UPDATE:

Thanks for the answers but I am looking for a proof using the rearrangement inequality in a similar way to the example.

Answer 1

Multiplying $s$ and $a_j$ by $n/s$ we can assume that $s=n$ and $a_1+\dotsm a_n=n$. Let us find the minimum of $$ \frac{a_1}{n-a_1}+\dots+\frac{a_n}{n-a_n} $$ over positive $a_j$ satisfying $a_1+\dotsm+a_n=n$. Using Lagrange multipliers, we take the function $$ g(a_1,\dots,a_n,\lambda)=\frac{a_1}{n-a_1}+\dots+\frac{a_n}{n-a_n}+\lambda(a_1+\dotsm+a_n-n) $$ and equate all of its derivatives with zero. We obtain $$ \frac{n}{(n-a_j)^2}-\lambda=0,\quad j=1,\dots,n,\qquad a_1+\dotsm+a_n=n. $$ It is easily seen that the only solution is $a_1=a_2=\dotsm=1$, $\lambda=n/(n-1)^2$.

The minimum is attained at this point and is $n/(n-1)$. This proves the desired assertion.

Answer 2

We need to prove that $\sum\limits_{i=1}^n\left(\frac{a_i}{s-a_i}-\frac{1}{n-1}\right)\geq0$ or $\sum\limits_{i=1}^n\frac{na_i-s}{s-a_i}\geq0$.

It's obvious that $\left(na_1-s,na_2-s,...,na_n-s\right)$ and $\left(\frac{1}{s-a_1},\frac{1}{s-a_2},...,\frac{1}{s-a_n}\right)$ are same ordered.

Thus, by Rearrangement (Chebyshov) we obtain:

$$n\sum\limits_{i=1}^n\frac{na_i-s}{s-a_i}\geq\sum\limits_{i=1}^n\left(na_i-s\right)\sum\limits_{i=1}^n\frac{1}{s-a_i}=0$$

Done!