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Let $x_{i}\ge 1$,show that $$\left(\sum_{i=1}^{n}x_{i}+n\right)^n\ge \left(\prod_{i=1}^{n}x_{i}\right)\left(\sum_{i=1}^{n}\dfrac{1}{x_{i}}+n\right)^n$$

or $$\left(\dfrac{\sum_{i=1}^{n}x_{i}+n}{\sum_{i=1}^{n}\dfrac{1}{x_{i}}+n}\right)^n\ge \prod_{i=1}^{n}x_{i}$$ and it seem use AM-GM inequality? $$\sum_{i=1}^{n}x_{i}\ge n\sqrt[n]{x_{1}x_{2}\cdots x_{n}}$$ $$\sum_{i=1}^{n}\dfrac{1}{x_{i}}\ge \dfrac{n}{\sqrt[n]{x_{1}x_{2}\cdots x_{n}}}$$ let $\sqrt[n]{x_{1}x_{2}\cdots x_{n}}=t$,since $$\Longleftrightarrow \left(\dfrac{t+1}{\frac{1}{t}+1}\right)^n\ge t^n$$But I can't it

Michael Rozenberg
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    So the "$+n$" term is part of the summations ? – Zubzub Nov 09 '17 at 07:17
  • Your last inequality holds (both sides are in fact equal), but it is not equivalent to your first inequality. – Martin R Nov 09 '17 at 08:53
  • Assuming the $n$ is outside the summation, one can make a little more progress toward the goal by applying AM-GM to the summation of $1+x_i$ as opposed to $x_i$. – John Barber Nov 09 '17 at 15:19
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    @AlexRavsky Do you think that Kantorovitch inequality can do the trick with some substitution ? https://en.wikipedia.org/wiki/Kantorovich_inequality –  Nov 11 '17 at 21:03
  • @FatsWallers No, I don't think so. – Alex Ravsky Nov 12 '17 at 06:08
  • @AlexRavsky I think to solve this problem we need to generalize it .See here for the case $n=3$ https://math.stackexchange.com/questions/2519854/a-problem-edf3abcdef3-geqabcdef-frac1a-frac1b-frac1 –  Nov 14 '17 at 17:38
  • /my old comment corrected/ Let $M(1)$ be arithmetic mean of $x_i$, $M(0)$ is their geometric mean, and $M(-1)$ is their harmonic mean. It can be easily checked that the inequality is equivalent to the inequality $(M(1)-M(0))M(-1)\ge M(0)-M(-1)$. – Alex Ravsky Nov 18 '17 at 18:19

2 Answers2

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We need to prove that $$\frac{\sum\limits_{k=1}^nx_k+n}{\sum\limits_{k=1}^n\frac{1}{x_k}+n}\geq\sqrt[n]{\prod\limits_{k=1}^nx_k}$$ and by the Vasc's EV Method (see here:

http://emis.ams.org/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf ,

Corollary 1.7(a))

it's enough to prove our inequality for $$x_1=b^n\leq x_2=...=x_n=a^n$$

where $a\geq1$ and $b\geq1$ or $$\frac{(n-1)a^n+b^n+n}{\frac{n-1}{a^n}+\frac{1}{b^n}+n}\geq a^{n-1}b$$ or $$ab^{n-1}\left((n-1)a^n-na^{n-1}b+b^n\right)\geq a^n-nab^{n-1}+(n-1)b^n.$$ Let $a=xb$.

Hence, $x\geq1$ and we need to prove that $$b^nx\left((n-1)x^n-nx^{n-1}+1\right)\geq x^n-nx+n-1$$ and since $b\geq1$ and by AM-GM $$(n-1)x^n-nx^{n-1}+1\geq0,$$ it's enough to prove that $$(n-1)x^{n+1}-nx^n+x\geq x^n-nx+n-1$$ or $f(x)\geq0$, where $$f(x)=(n-1)x^{n+1}-(n+1)x^n+(n+1)x-(n-1).$$ Now, $$f'(x)=(n+1)(n-1)x^n-n(n+1)x^{n-1}+n+1$$ and $$f''(x)=n(n+1)(n-1)x^{n-1}-(n-1)n(n+1)x^{n-2}\geq0.$$ Thus, $$f'(x)\geq f'(1)=0$$ and from here $$f(x)\geq f(1)=0$$ and we are done!

Michael Rozenberg
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we have to prove : $$\left(\dfrac{\sum_{i=1}^{n}x_{i}+n}{\sum_{i=1}^{n}\dfrac{1}{x_{i}}+n}\right)^n\ge \prod_{i=1}^{n}x_{i}$$

Let $x_i\geq 1$ be real numbers so we have : $$\frac{1}{\sum_{i=1}^{n}S_i}\left(\sum_{i=1}^{n}S_i[(-x_i+\frac{(\sum_{i=1}^{n}x_i)+n}{(\sum_{i=1}^{n}\frac{1}{x_i})+n})(n)x_i^{n-1}+x_i^n]\right)\ge \prod_{i=1}^{n}x_{i}$$

Where :

$$S_i=\frac{\prod_{i=1}^{n}x_{i}}{|((-x_i+\frac{(\sum_{i=1}^{n}x_i)+n}{(\sum_{i=1}^{n}\frac{1}{x_i})+n})(n)x_i^{n-1}+x_i^n)-1|}$$

And :

$$S_{min}=\frac{n\prod_{i=1}^{n}x_{i}}{|((-x_{min}+\frac{(\sum_{i=1}^{n}x_{i})+n}{(\sum_{i=1}^{n}\frac{1}{x_{i}})+n})(n)x_{min}^{n-1}+x_{min}^n)-1|}$$

After that we use the theorem 5 of this link :

You just have to replace the function by $\phi(x)=x^n$ with domain $[x_{min};x_{max}]$ and $d=\frac{(\sum_{i=1}^{n}x_i)+n}{(\sum_{i=1}^{n}\frac{1}{x_i})+n}$ and remark that we have : $x_{min}\leq d \leq x_{max}$ and put $S_i=W_i$ to get the LHS.

  • Your notation is confusing, because you use the same index $i$ in the nested sums in the second inequality. – Martin R Dec 04 '17 at 16:05
  • @Martin R Could you be more specific ? Thanks for your interest . –  Dec 07 '17 at 18:45
  • You write $\sum_{i=1}^n (... (\sum_{i=1}^n x_i) ...)$. The inner sum should use a different summation variable, e.g. $\sum_{j=1}^n x_j$. – Also I must admit that I cannot yet follow your argumentation. Is the first inequality supposed to be equivalent to the second one? – Martin R Dec 07 '17 at 19:17
  • Thanks I will correct it .It's not very clear like this... –  Dec 07 '17 at 19:24