[tex]\displaystyle Inegalitatea~initiala~este~echivalenta~cu \\ \\ \left( \frac{a}{b}+ \frac{b}{a} \right)+\left( \frac{b}{c}+ \frac{c}{b} \right)+\left( \frac{a}{c}+ \frac{c}{a} \right) \ge 6. \\ \\ Insa~se~cunoaste~inegalitatea~ \frac{x}{y}+ \frac{y}{x} \ge 2,~unde~x,y\ \textgreater \ 0. \\ \\ Conform~acestei~inegalitati,~fiecare~paranteza~este \ge 2,~si~deci \\ \\ inegalitatea~initiala~este~demonstrata.[/tex]
