cho x,y,t,z khac 0 cm: m=x/x+y+z +y/x+y+t +z/y+z+t +t/x+z+t co gt ko phai so tu nhien

Giải thích các bước giải:

 Do $x,y,z,t>0$

$\rightarrow \begin{cases}\dfrac{x}{x+y+z}>\dfrac{x}{x+y+z+t}\\\dfrac{y}{x+y+t}>\dfrac{y}{x+y+z+t}\\\dfrac{z}{y+z+t}>\dfrac{z}{x+y+z+t}\\\dfrac{t}{x+z+t}>\dfrac{t}{x+y+z+t}\end{cases}$

$\rightarrow \dfrac{x}{x+y+z}+\dfrac{y}{x+y+t}+\dfrac{z}{y+z+t}+\dfrac{t}{x+z+t}>\dfrac{x}{x+y+z+t}+\dfrac{y}{x+y+z+t}+\dfrac{z}{x+y+z+t}+\dfrac{t}{x+y+z+t}$

$\rightarrow \dfrac{x}{x+y+z}+\dfrac{y}{x+y+t}+\dfrac{z}{y+z+t}+\dfrac{t}{x+z+t}>\dfrac{x+y+z+t}{x+y+z+t}=1(1)$

Lại có:

$\begin{cases}\dfrac{x}{x+y+z}<\dfrac{x+t}{x+y+z+t}\\\dfrac{y}{x+y+t}<\dfrac{y+z}{x+y+z+t}\\\dfrac{z}{y+z+t}<\dfrac{z+x}{x+y+z+t}\\\dfrac{t}{x+z+t}<\dfrac{t+y}{x+y+z+t}\end{cases}$

$\rightarrow \dfrac{x}{x+y+z}+\dfrac{y}{x+y+t}+\dfrac{z}{y+z+t}+\dfrac{t}{x+z+t}<\dfrac{x+t}{x+y+z+t}+\dfrac{y+z}{x+y+z+t}+\dfrac{z+x}{x+y+z+t}+\dfrac{t+y}{x+y+z+t}$

$\rightarrow \dfrac{x}{x+y+z}+\dfrac{y}{x+y+t}+\dfrac{z}{y+z+t}+\dfrac{t}{x+z+t}<\dfrac{2(x+y+z+t)}{x+y+z+t}=2(2)$

$\rightarrow $Từ (1) và (2) $1<m<2$

$\rightarrow $m không là số tự nhiên