Given that $abc=1$, prove that $\frac{b}{c} + \frac{c}{a}+ \frac{a}{b} \ge \frac{1}{a} + \frac{1}{b} +\frac{1}{c}$.

Question

I have tried to do a bunch of different versions of AM-GM on this and the equivalent versions that come from substituting in $abc=1$ and I always flip the inequality by doing it.

Help is greatly appreciated.

Answer 1

Put $a = \dfrac{x}{y}, b = \dfrac{y}{z}, c = \dfrac{z}{x}$, then $x,y,z > 0$, and $abc = 1$, the inequality to prove is: $\dfrac{xy}{z^2}+\dfrac{yz}{x^2}+\dfrac{xz}{y^2} \ge \dfrac{x}{z}+\dfrac{z}{y}+\dfrac{y}{x}$, or equivalently:$(xy)^3+(yz)^3+(xz)^3 \ge (xy)^2(xz)+(yz)^2(xy)+(xz)^2(yz)$. Put $m = xy, n = yz, p = xz$, then we finally prove the inequality: $m^3+n^3+p^3 \ge m^2p+p^2n+n^2m$. By AM-GM inequality: $m^2p = m\cdot m\cdot p \le \dfrac{m^3+m^3+p^3}{3}, n^2m = n\cdot n\cdot m \le \dfrac{n^3+n^3+m^3}{3}, p^2n = p\cdot p\cdot n \le \dfrac{p^3+p^3+n^3}{3}$. Adding these $3$ inequalities we have the desire result.

Answer 2

Because for positive variables by AM-GM we obtain:

$\sum\limits_{cyc}\frac{a}{b}=\frac{1}{3}\sum\limits_{cyc}\left(\frac{a}{b}+\frac{2b}{c}\right)\geq\sum\limits_{cyc}\sqrt[3]{\frac{ab}{c^2}}=\sum\limits_{cyc}\frac{1}{c}$.

Done!