An inequality problem - Solution explanation

Question

I have recently been reading about inequalities, and in this section discussing AM-GM inequalities, I was faced by this question that I couldn't understand its solution. Note: I cant find a relationship between the solution and AM-GM. I would be very thankful if someone can take me through the solution.

The problem is as follows :-

Let $a_1,a_2,...,a_n$ be positive numbers such that $\frac{1}{1+a_1}+\frac{1}{1+a_2}+...+\frac{1}{1+a_n} = 1$. Prove that

$$\sqrt{a_1}+...+\sqrt{a_n}\ge(n-1)(\frac{1}{\sqrt{a_1}}+...+\frac{1}{\sqrt{a_n}})$$

The solution was :-

$\sum_{i=1}^n \frac{1}{1+a_i} = 1 \Rightarrow \sum_{i=1}^n \frac{a_i}{1+a_i} = n-1$

Observe that

$$\sum_{i=1}^n \sqrt{a_i}-(n-1)\sum_{i=1}^n \frac{1}{\sqrt{a_i}}=\sum_{i=1}^n \frac{1}{1+a_i}\sum_{i=1}^n \sqrt{a_i} - \sum_{i=1}^n \frac{a_i}{1+a_i}\sum_{i=1}^n \frac{1}{\sqrt{a_i}}$$

$$= \sum_{i,j} \frac{a_i-a_j}{(1+a_j)\sqrt{a_i}}=\sum_{i\gt j} \frac{(\sqrt{a_i}\sqrt{a_j}-1)(\sqrt{a_i}-\sqrt{a_j})^2(\sqrt{a_i}+\sqrt{a_j})}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}}$$

Since $1 \ge \frac{1}{1+a_i}+\frac{1}{1+a_j}=\frac{2+a_i+a_j}{1+a_i+a_j+a_ia_j}$, we can deduce that $a_ia_j \ge 1$. Hence the terms of the last sum are positive.

I am really struggling to understand this proof and if someone could simplify it and (maybe show where it is related to AM-GM inequalities), I would be very grateful.

Answer 1

The explanation is very dense and skips a lot of steps.

For the first implication, observe that $\sum_{i=1}^n \frac{a_i}{1+a_i} = \sum_{i=1}^n \frac{(1+a_i)-1}{1+a_i}=n-1$.

The first equality after "Observe that" is better written as: $$ \left(\sum_{i=1}^n \sqrt{a_i}\right)-(n-1)\left(\sum_{i=1}^n \frac{1}{\sqrt{a_i}}\right)= \left(\sum_{j=1}^n \frac{1}{1+a_j}\right)\left(\sum_{i=1}^n \frac{a_i}{\sqrt{a_i}}\right) - \left(\sum_{j=1}^n \frac{a_j}{1+a_j}\right)\left(\sum_{i=1}^n \frac{1}{\sqrt{a_i}}\right). $$ The second equality follows from an algebraic simplification of the RHS.

The final equality is trickier. We break the double sum over $i,j$ into three pieces: (a) $j>i$, (b) $j=i$, and (c) $j<i$. The (b) terms drop out. The (a) terms can be written $$\sum_i\sum_{j>i}\frac{a_i-a_j}{(1+a_j)\sqrt{a_i}} \stackrel{(*)}=\sum_j\sum_{i>j}\frac{a_j-a_i}{(1+a_i)\sqrt{a_j}} =-\sum_j\sum_{i>j}\frac{a_i-a_j}{(1+a_i)\sqrt{a_j}} $$ where in $(*)$ we swap the index $i$ with the index $j$. Combining this with the (c) terms we get $$\sum_{i>j}\left(\frac{a_i-a_j}{(1+a_j)\sqrt{a_i}}-\frac{a_i-a_j}{(1+a_i)\sqrt{a_j}}\right) $$ which, after some algebra, is the same as the final expression. The final deduction "$a_ja_j\ge1$" follows from rearranging the inequality $1 \ge\frac{2+a_i+a_j}{1+a_i+a_j+a_ia_j}$. Since each of the $a$'s is positive, it follows that $\sqrt{a_i}\sqrt{a_j}=\sqrt{a_ia_j}\ge1$. We deduce that every term in the final summation is a product of non-negative items, and the final summation is non-negative.

I don't see AM-GM anywhere in the proof.