SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Without loss of generality, assume that $a\ge b\ge c$. From the condition, we get $a\ge 1\ge c>0$. Write the inequality as $$\frac{a+b+c}{c}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\ge a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3$$ which is equivalent to $$(a+b+c-1)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1\right)\ge 4$$ Since $(a-1)(1-c)\ge 0$, we obtain $a+c\ge 1+ac$ and thus $\frac{1}{a}+\frac{1}{c}\ge 1+\frac{1}{ac}$. Therefore, $$\begin{array}{rl}& (a+b+c-1)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1\right)\\ & \ge (b+ac)\left(\frac{1}{b}+\frac{1}{ac}\right)=2+\frac{b}{ac}+\frac{ac}{b}\ge 2+2=4\end{array}$$ This completes the proof. Equality holds for $a=b=c=1$.