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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.

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  • $\begingroup$ You should be more precise in explaining what step you don't understand and what is clear to you. $\endgroup$
    – MrYouMath
    Commented Aug 19, 2016 at 16:50
  • $\begingroup$ @MrYouMath The proof is just a few steps and I do not understand how the answer develops from one step to the other. Also the end does not seem very clear to me (in terms of how it proves the statement above). $\endgroup$ Commented Aug 19, 2016 at 16:57

1 Answer 1

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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.

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  • $\begingroup$ Can you please explain how this proves the statement? $\endgroup$ Commented Aug 19, 2016 at 17:59
  • $\begingroup$ I've added more detail to tie up the proof. $\endgroup$
    – grand_chat
    Commented Aug 19, 2016 at 18:15

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