Question

Let x, y, z are distinct positive integers and

$m = (\frac{x^2 + y^2 + z^2}{x + y + z})^{(x+y+z)}$, $n = x^x y^y z^z$, $P = (\frac{x + y + z}{3})^{(x+y+z)}$, then

• A. m > n
• B. n > p
• C. m < n
• D. n < p

Answer 1

Answer: (A), (B)

(A), (B)

Answer 2

Let $S = x+y+z$.

$m = (\frac{x^2+y^2+z^2}{S})^S$, $n = x^x y^y z^z$, $P = (\frac{S}{3})^S$.

Since $x, y, z$ are distinct positive integers, $S > 0$.

Compare the bases of $m$ and $P$: $\frac{x^2+y^2+z^2}{S}$ and $\frac{S}{3}$.

The inequality $\frac{x^2+y^2+z^2}{x+y+z} > \frac{x+y+z}{3}$ holds because $(x-y)^2+(y-z)^2+(z-x)^2 > 0$ for distinct $x, y, z$.

Raising to the power $S$, we get $m > P$.

Use the weighted AM-GM inequality with values $x, y, z$ and weights $x, y, z$:

$\frac{x \cdot x + y \cdot y + z \cdot z}{x+y+z} > (x^x y^y z^z)^{\frac{1}{x+y+z}}$ for distinct $x, y, z$.

Raising to the power $S$, we get $m > n$.

Use Jensen's inequality for the convex function $f(t) = t \log t$ with weights $1, 1, 1$:

$\frac{x \log x + y \log y + z \log z}{3} \ge \frac{x+y+z}{3} \log(\frac{x+y+z}{3})$.

This simplifies to $(x^x y^y z^z)^{1/3} \ge (\frac{x+y+z}{3})^{(x+y+z)/3}$.

Raising to the power 3, we get $x^x y^y z^z \ge (\frac{x+y+z}{3})^{x+y+z}$.

This is $n \ge P$. Strict inequality holds for distinct $x, y, z$, so $n > P$.

Combining the results, $m > n > P$.

Therefore, $m > n$ and $n > P$.