If you let $ x= \log_m a; y = \log_m b$ and $ z = \log_m c$ the inequality becomes $ \frac{x}{y}+\frac{y}{z}+\frac{z}{x} \ge 3$ which is true by AM-GM

hsbhatt wrote: If you let $ x = \log_m a; y = \log_m b$ and $ z = \log_m c$ the inequality becomes $ \frac {x}{y} + \frac {y}{z} + \frac {z}{x} \ge 3$ which is true by AM-GM what's AM-GM?

The arithmetic mean is always greater than or equal to the geometric mean. In this case, $ \frac {\frac {x}{y} + \frac {y}{z} + \frac {z}{x}}{3} \ge \sqrt [3] {\frac {x}{y} \cdot \frac {y}{z} \cdot \frac {z}{x}} = 1$, hence $ \frac {x}{y} + \frac {y}{z} + \frac {z}{x} \ge 3$

one important part of this is that all of the numbers involved are nonnegative.

hsbhatt wrote: If you let $ x = \log_m a; y = \log_m b$ and $ z = \log_m c$ the inequality becomes $ \frac {x}{y} + \frac {y}{z} + \frac {z}{x} \ge 3$ which is true by AM-GM Doesn't it become... $ \log_c \dfrac{y}{z} + \log_b \dfrac{x}{y} +\log_a \dfrac{z}{x}$

How did you get $ \frac {x}{y} + \frac {y}{z} + \frac {z}{x} \ge 3$?

$ \frac{ln(log_cb)}{lnc} + \frac{ln(log_ba)}{lnb} + \frac{ln(log_ac)}{lna}\ge 0$ $ \frac {ln(lnb/lnc)}{lnc} + \frac {ln(lna/lnb)}{lnb} + \frac {ln(lnc/lna)}{lna}\ge 0$ $ \frac {lnlnb - lnlnc}{lnc} + \frac {lnlna - lnlnb}{lnb} + \frac {lnlnc - lnlna}{lna}\ge0$ Now, let $ a = e^x$, $ b = e^y$, $ c = e^z$ with $ x\gey\gez > 0$ $ \frac {lny - lnz}{z} + \frac {lnx - lny}{y} + \frac {lnz - lnx}{x} \ge 0$ $ \frac {lny}{z} + \frac {lnx}{y} + \frac {lnz}{x}\ge \frac {lnx}{x} + \frac {lny}{y} + \frac {lnz}{z}$ Which is rearrangement on oppositely sorted sequences $ lnx,lny,lnz$ and $ \frac {1}{x},\frac {1}{y},\frac {1}{z}$