[tex]\displaystyle Se~utilizeaza~inegalitatea~mediilor. \\ \\ Prezint~cateva~solutii: \\ \\ 1.~a+b+c+ab+bc+ac \ge 6\sqrt[6]{a \cdot b \cdot c \cdot ab \cdot bc \cdot ac}=6 \sqrt[6]{(abc^3)}=6. \\ \\ 2. \\ \\ a+bc \ge 2\sqrt{abc}=2. \\ \\ b+ac \ge 2 \sqrt{abc}=2. \\ \\ c+ab \ge 2 \sqrt{abc}=2. \\ \\ Prin~insumare~rezulta~concluzia. [/tex]
[tex]\displaystyle 3.~bc= \frac{1}{a}~;~ ac= \frac{1}{b}~;~ ab=\frac{1}{c}.~Inegalitatea~devine: \\ \\ a+\frac{1}{a}+ b+\frac{1}{b}+c+ \frac{1}{c} \ge 6,~adevarat,~caci~x+ \frac{1}{x} \ge 2 ~\forall~x\ \textgreater \ 0.[/tex]
