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Let $x_1,x_2,\ldots,x_n > 0$ such that $\dfrac{1}{1+x_1}+\cdots+\dfrac{1}{1+x_n}=1$. Prove the following inequality. $$\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}+\cdots+\dfrac{1}{\sqrt{x_n}} \right ).$$

This is Exercice 1.48 in the book "Inequalities-A mathematical Olympiad approach".

Attempt

I tried using HM-GM and I got $\left ( \dfrac{1}{x_1x_2\cdots x_n}\right)^{\frac{1}{2n}} \geq \dfrac{n}{\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}+\cdots+\dfrac{1}{\sqrt{x_n}}} \implies \dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_1}}+\cdots+\dfrac{1}{\sqrt{x_n}} \geq n(x_1x_2 \cdots x_n)^{\frac{1}{2n}}$. But I get stuck here and don't know if this even helps.

Arnaud D.
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Jacob Willis
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2 Answers2

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Note that $\displaystyle \sum_{i = 1}^n \dfrac{1}{1+a_i} = 1 \implies \sum_{i = 1}^n \dfrac{a_i}{1+a_i} = n-1$. Then, $$\displaystyle \sum_{i = 1}^n \sqrt{a_i}-(n-1)\sum_{i = 1}^n \dfrac{1}{\sqrt{a_i}} = \sum_{i = 1}^n \dfrac{1}{1+a_i}\sum \sqrt{a_i} - \sum_{i = 1}^n \dfrac{a_i}{1+a_i} \sum_{i = 1}^n \dfrac{1}{\sqrt{a_i}} = \displaystyle \sum_{i,j} \dfrac{a_i-a_j}{(1+a_j)\sqrt{a_i}} = \sum_{i>j} \dfrac{(\sqrt{a_i}\sqrt{a_j}-1)(\sqrt{a_i}-\sqrt{a_j})^2(\sqrt{a_i}+\sqrt{a_j})}{(1+a_i)(1+a_j)\sqrt{a_i}\sqrt{a_j}}$$

But $1 \geq \dfrac{1}{1+a_i}+\dfrac{1}{1+a_j} = \dfrac{2+a_i+a_j}{1+a_i+a_j+a_ia_j}$ thus $a_ia_j \geq 1$. Hence the terms of the last sum are positive.

Jacob Willis
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Rewrite our inequality in the following form: $\sum\limits_{i=1}^n\frac{x_i+1}{\sqrt{x_i}}\sum\limits_{i=1}^n\frac{1}{x_i+1}\geq n\sum\limits_{i=1}^n\frac{1}{\sqrt{x_i}}$, which is Chebyshov.

Michael Rozenberg
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  • But to use Chebyshev, you need the terms to be decreasing, right? It's not necessarily true. – Vinícius Novelli Dec 29 '15 at 22:07
  • To Vinícius Novelli. If $x_1\leq x_2\leq...\leq x_n$ so $x_2\geq\frac{1}{x_1}>1$ – Michael Rozenberg Dec 29 '15 at 22:25
  • It was for $x_1<1$. But If $x_1\geq1$ it's obvious. – Michael Rozenberg Dec 29 '15 at 22:34
  • $\frac{x+1}{\sqrt{x}}$ decreases for $x<1$. So the Chebyshev does not apply directly. – Hans Oct 12 '18 at 01:37
  • What do you think of my comment above? – Hans Dec 14 '19 at 09:36
  • @Hans See please my previous comments. – Michael Rozenberg Dec 14 '19 at 15:36
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    Ingenious proof! +1 – Hans Dec 17 '19 at 04:27
  • @MichaelRozenberg I know this answer is old, but I posted a question (4807666) and the answer redirected to this topic. Your proof seems incorrect. I tried in the same way proving my inequality and I encountered a problem. Your inequality is false. Fore example n=2, $x_1=0.1$, $x_2=1$. These numbers don't satisfy the condition but your proof doesn't use it in proving your inequality. – Mateo Nov 16 '23 at 02:19
  • @MichaelRozenberg, now I see you're right. Namely, if $x_1<1$ then $(x_1+1)/\sqrt{x_1}\leq (x_2+1)/\sqrt{x_2}$, although the function isn't monotone. I think this observation could be placed in your answer. Thank you for your solution. – Mateo Nov 16 '23 at 02:46