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 $x_1=x_2=\dots=x_n$.
$\blacksquare\;\;\;\blacksquare$
