[tex]Notez \ \sqrt{3a+1}=x, \ \sqrt{3b+1}=y, \ \sqrt{3c+1}=z\\ \\ \frac{3}{\frac{1}{\sqrt{3a+1}}+\frac{1}{\sqrt{3b+1}}+\frac{1}{\sqrt{3c+1}}}\leq \sqrt[3]{\sqrt{3a+1}\cdot \sqrt{3b+1} \cdot \sqrt{3c+1}}\leq \\ \\ \leq \frac{\sqrt{3a+1}+\sqrt{3b+1}+\sqrt{3c+1}}{3}\leq \sqrt{\frac{(\sqrt{3a+1})^2+(\sqrt{3b+1})^2+(\sqrt{3c+1})^2}{3}}\\ \\ Noi \ avem \ din \ enunt \ proprietatea \ ca \ a+b+c=3, \ ceea \ ce \ inseamna \ ca \\ \\vom \ ajunge \ la \ un \ moment \ dat \ sa \ o \ folosim \ in \ membrul \ drept\\ \\ Mergem \ pe \ x+y+z=S\leq ...\\ \\ Nu \ avem \ nicio \ informatie \ despre \ produsele \ ab, \ bc, \ ac \\ \\ \Rightarrow nu \ ne \ va fi \ utila \ media \ geometrica \ si \ nici \ cea \ armonica\\ \\ E \ evident \ ca \ vom \ folosi \ media \ aritmetica, \ m_a=\frac{S}{3}, \\ \\ iar \ media \ patratica \ deasemenea \ ne \ va fi \ utila: \\ \\ OBS \ ca: \\ \\ (\sqrt{3a+1})^2=|3a+1|=3a+1, \ (3a+1\geq 0, \ a\geq 0)\\ \\ \\ \\ \frac{\sqrt{3a+1}+\sqrt{3b+1}+\sqrt{3c+1}}{3}\leq \sqrt{\frac{(\sqrt{3a+1})^2+(\sqrt{3b+1})^2+(\sqrt{3c+1})^2}{3}}\\ \\ \Rightarrow \frac{S}{3}\leq \sqrt{\frac{|3a+1|+|3b+1|+|3c+1|}{3}}\\ \\ \Rightarrow \frac{S}{3}\leq \sqrt{\frac{3a+1+3b+1+3c+1}{3}}\\ \\ \Rightarrow \frac{S}{3}\leq \sqrt{\frac{3(a+b+c+1}{3}}\\ \\ \Rightarrow \frac{S}{3}\leq \sqrt{(a+b+c)+1}\\ \\ \Rightarrow \frac{S}{3}\leq \sqrt{3+1}\\ \\ \Rightarrow S\leq 3\cdot 2\\ \\ \Rightarrow \boxed{S\leq 6}[/tex]
