Am V Ă Z U T - O ... ! N-am apucat să-i scriu... // A problem solved as a consolation

 Ieri a fost pe chat. Am văzut-o când am intrat, noaptea devreme la  [0:30]. Avea "+voice", semn că a dat "!like"... Dar deja nu mai era activă, probabil a rămas doar conectată cu laptopul...

Cu IP ul tradițional :   sanda@WwW.ChatApropo.Ro/-188926E.hosted-by-worldstram.net

Iar EU am rămas pe Whtspp - BLOCAT - și pe Blog  - PĂRĂSIT

(poate nu abandonat de tot, dar de la 10 000 de vizualizări am scăzut brusc în ultimele zile)

(Este ciudat acest  www.aocr.org  care a deținut recordul de accesări pe ultimele 15 zile)

     Let us, however, console ourselves with the following Problem (page 140 - rom ; page 143 - engl) : 

"O.G : 288.  If  $x_1,\;x_2,\dots,\;x_n\;>0\;,\;\;n\in\mathbb{N}\;,\;n\geqslant 2$  and

                                 $S=x_1+x_2+\dots+x_n$  , show that

$\frac{x_1}{S-x_1}+\frac{x_2}{S-x_2}+\dots +\frac{x_n}{S-x_n}\geqslant \frac{n}{n-1}\;.$

 {No author : } * * *"

ANSWER CiP

$"="\;if\;and\;only\;if\;\;x_1=x_2=\dots=x_n$

                                Solution CiP

               For  $a_i\;>0$ , we have the inequality

$$\left (\sum_{i=1}^na_i\right )\cdot \left (\sum_{i=1}^n\frac{1}{a_i} \right )\geqslant n^2 \tag{1}$$

with  $"="\Leftrightarrow a_1=a_2=\dots=a_n.$ There are countless proofs for  (1)  . We provide one here via calculation, relying solely on

$\frac{a_i}{a_j}+\frac{a_j}{a_i}\geqslant 2\;\;,\;"="\Leftrightarrow a_i=a_j \tag{2}$

$(a_i-a_j)^2\geqslant 0\Rightarrow a_i^2+a_j^2\geqslant  2a_ia_j\Rightarrow \frac{a_i^2+a_j^2}{a_ia_j}\geqslant 2\;etc.$

      The left-hand side of  (1)  is

$$L:=\left (\sum_{i=1}^na_i\right )\cdot \left (\sum_{j=1}^n\frac{1}{a_j}\right )=\sum_{i=j=1}^n\frac{a_i}{a_j}+\sum_{1\leqslant i< j\leqslant n}\left (\frac{a_i}{a_j}+\frac{a_j}{a_i}\right ) \tag{3}$$

The number of pairs  $(i, j)$  with property  $1\leqslant i<j\leqslant n$  is  $\frac{(n-1)n}{2}$  according to the table :

$i=1\;\;,\;j=2,\;3,\;\dots,\;n\;\;||(n-1)\;pairs$

$i=2\;\;,\;\;\;\;\;\;j=3,\dots,\;n\;\;||(n-2)\;pairs$

$i=3\;\;,\;\;\;\;\;\;j=4,\dots,\;n\;\;||(n-3)\;pairs$

.....................................................................................

$i=n-2\;\;,\;j=n-1,\;n\;\;||2\;pairs$

$i=n-1\;\;,\;\;\;\;\;\;\;\;\;\;\;\;j=n\;\;||1\;pair$

so a total of  $1+2+\dots+(n-1)=\frac{(n-1)n}{2}$  pairs.

     Applying  (2)  to  (3)  $\frac{(n-1)n}{2}\;times$ , we obtain

$L\geqslant n+2\cdot \frac{(n-1)n}{2}=n+(n^2-n)=n^2$

that is  (1).

          If in the inequality  (1)  we set  $a_i=S-x_i\;, i=1,\;2,\;\dots,n$  we obtain

 $$(n-1)S\cdot \left (\sum_{i=1}^n\frac{1}{S-x_i}\right )\geqslant n^2\Rightarrow \sum_{i=1}^n\frac{S}{S-x_i}\geqslant \frac{n^2}{n-1}\Rightarrow$$

$$\Rightarrow \sum_{i=1}^n\frac{(S-x_i)+x_i}{S-x_i}\geqslant \frac{n^2}{n-1}\Rightarrow n+\sum_{i=1}^n\frac{x_i}{S-x_i}\geqslant \frac{n^2}{n-1}\Rightarrow$$

$$\Rightarrow \sum_{i=1}^n\frac{x_i}{S-x_i}\geqslant \frac{n^2}{n-1}-n=\frac{n}{n-1}.$$

We have proven the inequality stated in the problem. The  "="  sign holds if and only if  $S-x_1=S-x_2=\dots=S-x_n$  so  $x_1=x_2=\dots=x_n$.

$\blacksquare\;\;\;\blacksquare$