Question
Let $x,y,z$ are distinct positive integers and $m = \left( \frac{x^{2} + y^{2} + z^{2}}{x + y + z} \right)^{(x + y + z)},n = x^{x}y^{y}z^{z},P = \left( \frac{x + y + z}{3} \right)^{(x + y + z)}$, then
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Solution
Sol. [Key Concept: Use the divisibility / ordering relations among the integers.]
$$\frac{\underset{x\text{~}\text{times}\text{~}}{\overset{(x + x + \ldots.x)}{︸}} + \underset{y\text{~}\text{times}\text{~}}{\overset{(y + y + \ldots.y)}{︸}} + \underset{z\text{~}\text{times}\text{~}}{\overset{(z + z + \ldots.z)}{︸}}}{(x + y + z)} \geq \left( x^{x}y^{y}z^{z} \right)\frac{1}{x + y + z}$$
$$\geq \frac{x + y + z}{\underset{x\text{~}\text{times}\text{~}}{\overset{\left. \ \frac{1}{x} + \frac{1}{x} + \ldots \cdot \frac{1}{x} \right)}{︸}} + \underset{y\text{~}\text{times}\text{~}}{\overset{\left( \frac{1}{y} + \frac{1}{y} + \ldots \cdot \frac{1}{y} \right)}{︸}} + \underset{z\text{~}\text{times}\text{~}}{\overset{\left( \frac{1}{z} + \frac{1}{z} + \ldots \cdot + \frac{1}{z} \right)}{︸}}}$$
✔ (a) $mge\ n$ by the power-mean inequality
✔ (b) $nge\ p$ by weighted AM–GM
✘ (c) ordering is $mge\ nge\ p$
✘ (d) ordering is $mge\ nge\ p$
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