Cho x, y, z, t Î ℕ^*. Chứng minh rằng:

\[M = \frac{x}{{x + y + z}} + \frac{y}{{x + y + t}} + \frac{z}{{y + z + t}} + \frac{t}{{x + z + t}}\] không phải số tự nhiên.

\[M = \frac{x}{{x + y + z}} + \frac{y}{{x + y + t}} + \frac{z}{{y + z + t}} + \frac{t}{{x + z + t}}\]

Ta có:

\[\frac{x}{{x + y + z}} > \frac{x}{{x + y + z + t}}\]

\[\frac{y}{{x + y + t}} > \frac{y}{{x + y + z + t}}\]

\[\frac{z}{{y + z + t}} > \frac{z}{{x + y + z + t}}\]

\[\frac{t}{{x + z + t}} > \frac{t}{{x + y + z + t}}\]

\[ \Rightarrow M > \frac{x}{{x + y + z + t}} + \frac{y}{{x + y + z + t}} + \frac{z}{{x + y + z + t}} + \frac{t}{{x + y + z + t}}\]

Þ M > 1

Lại có:

\[\frac{x}{{x + y + z}} < \frac{{x + t}}{{x + y + z + t}}\]

\[\frac{y}{{x + y + t}} < \frac{{y + z}}{{x + y + z + t}}\]

\[\frac{z}{{y + z + t}} < \frac{{z + x}}{{x + y + z + t}}\]

\[\frac{t}{{x + z + t}} < \frac{{t + y}}{{x + y + z + t}}\]

\[ \Rightarrow M < \frac{{x + t}}{{x + y + z + t}} + \frac{{y + z}}{{x + y + z + t}} + \frac{{z + x}}{{x + y + z + t}} + \frac{{t + y}}{{x + y + z + t}}\]

\[ \Rightarrow M < \frac{{2(x + y + z + t)}}{{x + y + z + t}}\]

Þ M < 2

Do đó, 1 < M < 2

Vậy M không phải là số tự nhiên.