# $a^{x}\left( y-z\right) +a^{y}\left( z-x\right) +a^{z}\left( x-y\right) >0$

If a is any positive number except $1$ , and $x, y, z,$ are REALS no two of which are equal, then

$a^{x}\left( y-z\right) +a^{y}\left( z-x\right) +a^{z}\left( x-y\right) >0$.

It is quite easy to prove the case when $x, y, z$ are RATIONALS:

assuming WLOG that $x>y>z,$ and considering $x, y, z$ as integers, we apply the AM-GM inequality to the set of numbers $a^{x}, a^{x}, a^{x},…, a^{z}, a^{z}, a^{z}, …,$ where $a^{x}$ occurs$\left( y-z\right)$ times and $a^{z}$ occurs $\left( x-y\right)$ times; then, considering $x, y, z$ as fractions, and denoting them by $p/d, q/d, r/d$ where $p, q, r, d$ are integers and $d$ is positive, we can easily deduce this case from the preceding.

But how to prove this inequality in the real case (and not using the calculus methods)?

Use weighted AM-GM in exactly the same fashion. WLOG $x>y>z$, and
$$a^x(y-z)+a^z(x-y) > (x-z)a^{\frac{x(y-z)}{x-z}}a^{\frac{z(x-y)}{x-z}}=(x-z)a^y$$