How prove $ \sum (a+b+c+d)\sum\frac{ab+ac+ad+bc+bd+cd}{a+b+c+d}\sum\frac{abc+abd+bcd+cda}{ab+ac+ad+bc+bd+cd}\leq \sum a\sum b\sum c\sum d$

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.

Solution 1:

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

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