Question

$Daca~x,y,z-apartin-(0,inf),dem: \\ \frac{xy}{xy+x+y} + \frac{yz}{yz+y+z} + \frac{zx}{zx+z+x} \leq \frac{6+ x^{2} + y^{2}+ z^{2} }{9}$

Answer 1

[tex]\displaystyle\frac{1}{1+\frac{1}{x}+\frac{1}{y}}\leq \frac{1}{18}(4+x^2+y^2)\ \textless \ =\ \textgreater \ \\ (1+\frac{1}{x}+\frac{1}{y})(4+x^2+y^2)\geq 18\\ f(x,y)=(1+\frac{1}{x}+\frac{1}{y})(4+x^2+y^2)\\ (x_0,y_0)=(1,1)\ punct\ de\ minim\ minf(1,1)=18\\ \ \frac{1}{1+\frac{1}{z}+\frac{1}{y}}\leq \frac{1}{18}(4+z^2+y^2)\\ \frac{1}{1+\frac{1}{x}+\frac{1}{z}}\leq \frac{1}{18}(4+x^2+z^2)\\ Insumand\ cele \ 3\ inegalitati\\ rezulta\ inegalitatea\ din \ enunt.[/tex]

Answer 2

$\displaystyle Vom~"sparge"~inegalitatea,~dupa~cum~urmeaza. \\ \\ Vom~cauta~o~inegalitate~pentru~fiecare~termen,~astfel~ca,~prin \\ \\ insumarea~acestor~inegalitati~sa~rezulte~concluzia. \\ \\ \frac{6+x^2+y^2+z^2}{9}= \sum \frac{4+x^2+y^2}{18} ~(pana~acum,~tot~ceea~ce~am \\ \\ facut~a~fost~sa~caut~o~exprimare~simetrica~in~functie~de~(x,y), \\ \\ (y,z)~si~(x,z)~pentru~membrul~drept.) \\ \\ ------------------------------ \\ \\ Avem~de~demonstrat~ca~\sum \frac{xy}{xy+x+y} \leq \sum \frac{4+x^2+y^2}{18}.$

$\displaystyle~Deci~este~suficient~sa~demonstram~ca~\frac{xy}{xy+x+y} \leq \frac{4+x^2+y^2}{18}, \\ \\ adica~(xy+x+y)(4+x^2+y^2) \ge 18xy. \\ \\ Demonstratia~propriu-zisa~incepe~acum. \\ \\ Din~inegalitatea~mediilor,~avem~ \\ \\ xy+x+y \geq 3 \sqrt[3]{xy \cdot x \cdot y}=3 \sqrt[3]{x^2y^2}..........(1) \\ \\ Tot~din~inegalitatea~mediilor~avem~succesiv: \\ \\ 4+x^2+y^2 \geq 4+2xy=2(1+1+xy) \geq 2 \cdot 3 \sqrt[3]{1 \cdot 1 \cdot xy}=6 \sqrt[3]{xy}.....(2)$

$\displaystyle Prin~inmultirea~relatiilor~(1)~si~(2)~rezulta~concluzia.$