[tex]\displaystyle Vom~"sparge"~inegalitatea,~dupa~cum~urmeaza. \\ \\ Vom~cauta~o~inegalitate~pentru~fiecare~termen,~astfel~ca,~prin \\ \\ insumarea~acestor~inegalitati~sa~rezulte~concluzia. \\ \\ \frac{6+x^2+y^2+z^2}{9}= \sum \frac{4+x^2+y^2}{18} ~(pana~acum,~tot~ceea~ce~am \\ \\ facut~a~fost~sa~caut~o~exprimare~simetrica~in~functie~de~(x,y), \\ \\ (y,z)~si~(x,z)~pentru~membrul~drept.) \\ \\ ------------------------------ \\ \\ Avem~de~demonstrat~ca~\sum \frac{xy}{xy+x+y} \leq \sum \frac{4+x^2+y^2}{18}.[/tex]
[tex]\displaystyle~Deci~este~suficient~sa~demonstram~ca~\frac{xy}{xy+x+y} \leq \frac{4+x^2+y^2}{18}, \\ \\ adica~(xy+x+y)(4+x^2+y^2) \ge 18xy. \\ \\ Demonstratia~propriu-zisa~incepe~acum. \\ \\ Din~inegalitatea~mediilor,~avem~ \\ \\ xy+x+y \geq 3 \sqrt[3]{xy \cdot x \cdot y}=3 \sqrt[3]{x^2y^2}..........(1) \\ \\ Tot~din~inegalitatea~mediilor~avem~succesiv: \\ \\ 4+x^2+y^2 \geq 4+2xy=2(1+1+xy) \geq 2 \cdot 3 \sqrt[3]{1 \cdot 1 \cdot xy}=6 \sqrt[3]{xy}.....(2)[/tex]
[tex]\displaystyle Prin~inmultirea~relatiilor~(1)~si~(2)~rezulta~concluzia.[/tex]
