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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. $$\sqrt{x_1}+\sqrt{x_2}+\cdots+\sqrt{x_n} \geq (n-1) \left (\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}+\cdots+\dfrac{1}{\sqrt{x_n}} \right ).$$

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.

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    $\begingroup$ from where does this come? $\endgroup$ – Dr. Sonnhard Graubner Dec 29 '15 at 19:46
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    $\begingroup$ An inequality book. $\endgroup$ – Jacob Willis Dec 29 '15 at 19:56
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    $\begingroup$ Would you mind to tell us title and author? Does that book contain solutions or hints? – Or is this meant as an exercise for us? $\endgroup$ – Martin R Dec 29 '15 at 20:00
  • $\begingroup$ It was a problem my instructor gave me, and I don't know where he got it from. We must figure this out ourselves. $\endgroup$ – Jacob Willis Dec 29 '15 at 20:11
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    $\begingroup$ "We must figure this out for ourselves" - I have voted to close. $\endgroup$ – Carl Mummert Mar 7 '18 at 18:25
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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.

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  • $\begingroup$ Wow! Very nice proof! $\endgroup$ – Michael Rozenberg Dec 29 '15 at 22:40
  • $\begingroup$ +1. Very nice indeed! $\endgroup$ – Hans Oct 12 '18 at 1:40
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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.

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  • $\begingroup$ But to use Chebyshev, you need the terms to be decreasing, right? It's not necessarily true. $\endgroup$ – Vinícius Novelli Dec 29 '15 at 22:07
  • $\begingroup$ To Vinícius Novelli. If $x_1\leq x_2\leq...\leq x_n$ so $x_2\geq\frac{1}{x_1}>1$ $\endgroup$ – Michael Rozenberg Dec 29 '15 at 22:25
  • $\begingroup$ It was for $x_1<1$. But If $x_1\geq1$ it's obvious. $\endgroup$ – Michael Rozenberg Dec 29 '15 at 22:34
  • $\begingroup$ $\frac{x+1}{\sqrt{x}}$ decreases for $x<1$. So the Chebyshev does not apply directly. $\endgroup$ – Hans Oct 12 '18 at 1:37

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