$$ \frac ab + \frac bc + \frac ca \geq a + b + c $$ where $abc = 1$
Using AM-GM I can get $$ \frac ab + \frac bc +\frac ca + ab + ac + bc \geq 2(a + b + c), $$ but I can't show $ab + bc + ac \leq a + b + c$ because it isn't true, nor is it with the inequality reversed. If the above inequality is true doesn't $ab + ac + bc \leq a + b +c$ need to be true?
Using the substitution $a=\frac xy$, $b=\frac yz$ and $c=\frac zx$, we want to prove that $$ \sum_{cyc}\frac {xy}{z^2}\geq \frac xy+\frac yz+\frac zx $$ where we are taking cyclic sums. Using AM-GM, we find that $$ \frac{\frac{xy}{z^2}+\frac{xz}{y^2} +\frac{xz}{y^2}}3 \geq \sqrt[3]{\frac{x^3}{y^3}}=\frac xy $$ Applying this three times yields the required inequality.