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let $a_{i}>0,b_{i}>0,c_{i}>0,d_{i}>0,i=1,2,\cdots,n $

show that $$\sum_{i=1}^{n}(a_{i}+b_{i}+c_{i}+d_{i})\sum_{i=1}^{n}\dfrac{a_{i}b_{i}+b_{i}c_{i}+c_{i}d_{i}+d_{i}a_{i}+a_{i}c_{i}+b_{i}d_{i}}{a_{i}+b_{i}+c_{i}+d_{i}}\sum_{i=1}^{n}\dfrac{a_{i}b_{i}c_{i}+b_{i}c_{i}d_{i}+c_{i}d_{i}a_{i}+d_{i}a_{i}b_{i}}{b_{i}c_{i}+c_{i}a_{i}+a_{i}b_{i}+d_{i}a_{i}+d_{i}b_{i}+d_{i}c_{i}}\sum_{i=1}^{n}\dfrac{a_{i}b_{i}c_{i}d_{i}}{a_{i}b_{i}c_{i}+b_{i}c_{i}d_{i}+c_{i}d_{i}a_{i}+d_{i}a_{i}b_{i}}\le\sum_{i=1}^{n}a_{i}\sum_{i=1}^{n}b_{i}\sum_{i=1}^{n}c_{i}\sum_{i=1}^{n}d_{i}$$

I think this true, because yesterday I have proved this

$\sum a+b+c \sum \frac{ab+bc+ca}{a+b+c} \sum \frac{abc}{ab+bc+ca} \leq \sum a \sum b \sum c$

Thank you everyone.

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math110
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    Wow. Assuming I was drunk enough to try proving this monstrosity-I'd try induction. – Mathemagician1234 Nov 04 '13 at 08:13
  • Thank you,can you post your solution? – math110 Nov 04 '13 at 12:03
  • I'll try and get it done today,but rather busy. – Mathemagician1234 Nov 04 '13 at 20:01
  • I'm halfway through it and it's just an ugly computation. just ugly.Not difficult,but just grotesquely tedious.Wish I still had my MATHEMATICA right now,would make it a lot quicker since I could get it to do the spade work. – Mathemagician1234 Nov 05 '13 at 04:21
  • I'm exhausted. If I have time before the deadline,I'll try and finish it.I got stuck on how to rearrange it correctly for the inductive step. I'm VERY sure this how to do it,but too tired to work it out now and can't drink coffee for health reasons. I hate being sick........... – Mathemagician1234 Nov 05 '13 at 08:11
  • I'm curious if $S_{k+1}/S_k$ is always a concave function of its arguments where $S_m$ is the $m$-th symmetric sum. This would give the desired result in no time. For $3$ variables it is, indeed, so, and for $4$ variables only $S_3/S_2$ is unclear. – fedja Aug 15 '16 at 00:20
  • In all honesty I don't even understand the question. Without parentheses, I don't know which sigmas are inside of which other sigmas, and what is being multiplied by what. – A. Thomas Yerger Oct 10 '16 at 21:31

1 Answers1

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$$\sum a_i \sum bi \geq \sum ( a_i +b_i ) \sum \frac{a_i b_i}{a_i+b_i}( Milne)$$
$$\sum c_i \sum di \geq \sum ( c_i +d_i ) \sum \frac{c_i d_i}{c_i+d_i}( Milne) $$ $$ \sum a_i \sum bi \sum c_i \sum di \geq \sum ( a_i +b_i ) \sum \frac{a_i b_i}{a_i+b_i} \sum ( c_i +d_i ) \sum \frac{c_i d_i}{c_i+d_i}$$ $$ = \sum ( a_i +b_i ) \sum ( c_i +d_i )\sum \frac{a_i b_i}{a_i+b_i} \sum \frac{c_i d_i}{c_i+d_i} $$ $$ (Milne ) \geq \sum ( a_i +b_i + c_i+d_i )\sum \frac{(a_i+ b_i)(c_i+d_i)}{a_i+b_i+c_i+d_i} \sum(\frac{a_i b_i}{a_i+b_i}+\frac{c_i d_i}{c_i+d_i})\sum \frac{\frac{a_i b_i }{a_i+b_i}\frac{c_i d_i}{c_i+d_i}}{\frac{a_i b_i }{a_i+b_i}+\frac{c_i d_i}{c_i+d_i}} $$ $$ \geq \sum ( a_i +b_i + c_i+d_i )\sum \frac{(a_i+ b_i)(c_i+d_i)}{a_i+b_i+c_i+d_i} \sum \frac{a_ib_i(c_i+d_i)+c_id_i(a_i+b_i)}{(a_i+b_i)(c_i+d_i)} \sum \frac{a_ib_ic_id_i}{a_ib_i(c_i+d_i)+c_id_i(a_i+b_i)}$$

Ref : Dan Sitaru from "Imad Zak Math Group" on facebook

mounir ben salem
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