Lời giải:
$\quad M = \dfrac{x}{x+y+z} +\dfrac{y}{y+z+t} +\dfrac{z}{z+t+x}+\dfrac{t}{t+x+y}$
Ta có:
$\dfrac{x}{x+y+z}>\dfrac{x}{x+y+z+t}$
$\dfrac{y}{y+z+t}>\dfrac{y}{x+y+z+t}$
$\dfrac{z}{z+t+x}>\dfrac{z}{x+y+z+t}$
$\dfrac{t}{t+x+y}>\dfrac{t}{x+y+z+t}$
Cộng vế theo vế ta được:
$\dfrac{x}{x+y+z} +\dfrac{y}{y+z+t} +\dfrac{z}{z+t+x}+\dfrac{t}{t+x+y} >\dfrac{x+y+z+t}{x+y+z+t}$
$\Leftrightarrow M > 1\qquad (1)$
Ta lại có:
$\dfrac{x}{x+y+z}<\dfrac{x}{x+z}$
$\dfrac{y}{y+z+t}<\dfrac{y}{y+t}$
$\dfrac{z}{z+t+x}<\dfrac{z}{x+z}$
$\dfrac{t}{t+x+y}<\dfrac{t}{y+t}$
Cộng vế theo vế ta được:
$\dfrac{x}{x+y+z} +\dfrac{y}{y+z+t} +\dfrac{z}{z+t+x}+\dfrac{t}{t+x+y} < \dfrac{x}{x+z} +\dfrac{z}{x+z} +\dfrac{y}{y+t} +\dfrac{t}{y+t}$
$\Leftrightarrow M < 2\qquad (2)$
Từ $(1)(2)\Rightarrow 1 < M < 2$
$\Rightarrow M$ không phải số tự nhiên
