Ta có: A = \(\dfrac{x}{x+y}\) + \(\dfrac{y}{y+z}\) + \(\dfrac{z}{z+x}\)
\(\dfrac{x}{x+y+z}\) < \(\dfrac{x}{x+y}\)
\(\dfrac{y}{x+y+z}\) < \(\dfrac{y}{y+z}\)
\(\dfrac{z}{x+y+z}\) < \(\dfrac{z}{z+x}\)
Do đó \(\dfrac{x+y+z}{x+y+z}\) < A
1 < A (1)
Vì x;y;z > 0 (x;y;z nguyên dương) \(\Rightarrow\) x < x + y
xz < (x + y)z
xz + (x + y)z < (x + y)z + (x + y)x
x(x + y + z) < (x + y)(x+ z)
\(\Rightarrow\) \(\dfrac{x}{x+y}\) < \(\dfrac{x+z}{x+y+z}\)
Tương tự: \(\dfrac{y}{y+z}\) < \(\dfrac{y+x}{x+y+z}\)
\(\dfrac{z}{z+x}\) < \(\dfrac{z+y}{x+y+z}\)
Hay A < \(\dfrac{2\left(x+y+z\right)}{x+y+z}\)
A < 2 (2)
Từ (1) và (2) nên 1 < A < 2.
Vì 1 và 2 là hai số tự nhiên liên tiếp nên A không phải là số nguyên.
Ta có:
\(\dfrac{x}{x+y}>\dfrac{x}{x+y+z}\)
\(\dfrac{y}{z+y}>\dfrac{y}{x+y+z}\)
\(\dfrac{z}{x+z}>\dfrac{z}{x+y+z}\)
\(\Rightarrow A=\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}>\dfrac{x+y+z}{x+y+z}=1\)\(\Rightarrow A>1\left(1\right)\)
Lại có
\(\dfrac{x}{x+y}< \dfrac{x+z}{x+y+z}\)
\(\dfrac{y}{z+y}< \dfrac{y+z}{x+y+z}\)
\(\dfrac{z}{x+z}< \dfrac{z+y}{x+y+z}\)
\(\Rightarrow A=\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}< \dfrac{2\left(x+y+x\right)}{x+y+z}=2\)\(\Rightarrow A< 2\left(2\right)\)
Từ (1),(2)\(\Rightarrow1< A< 2\) \(\RightarrowĐPCM\)
Đúng rồi đó Quang, mẹ ta có bày rồi
