* C/m : A > 1

Ta có :

\(\frac{x}{x+y}>\frac{x}{x+y+z}\)( vì x > 0 ; 0 < x + y < x + y + z )

\(\frac{y}{y+z}>\frac{y}{x+y+z}\)( vì y > 0 ; 0 < y + z < x + y + z )

\(\frac{z}{z+x}>\frac{z}{x+y+z}\)( vì z > 0 ; 0 < z + x < x + y + z )

\(\Rightarrow\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}>\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}\)

\(\Rightarrow A>\frac{x+y+z}{x+y+z}\Rightarrow A>1\)

* C/m : A < 2 

Áp dụng BĐT : \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+n}{b+n}\) ( a,b,n \(\in\)N* )

Với x,y,z \(\in\)N* ta có :

- Vì : 0 < x < x + y \(\Rightarrow\frac{x}{x+y}< 1\Rightarrow\frac{x}{x+y}< \frac{x+z}{x+y+z}\)

- Vì : 0 < y < y + z \(\Rightarrow\frac{y}{y+z}< 1\Rightarrow\frac{y}{y+z}< \frac{x+y}{x+y+z}\)

- Vì : 0 < z < z + x \(\Rightarrow\frac{z}{z+x}< 1\Rightarrow\frac{z}{z+x}< \frac{z+y}{x+y+z}\)

\(\Rightarrow\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}< \frac{x+z}{x+y+z}+\frac{x+y}{x+y+z}+\frac{y+z}{x+y+z}\)

\(\Rightarrow A< \frac{x+z+x+y+y+z}{x+y+z}\Rightarrow A< \frac{2\left(x+y+z\right)}{x+y+z}\Rightarrow A< 2\)

Mà A < 1 => 1 < A < 2 ; 1 và 2 là hai số nguyên liên tiếp

=> A không có giá trị nguyên 

Vậy ...

Ta có: \(\frac{x}{x+y}>\frac{x}{x+y+z}\)

\(\frac{y}{y+z}>\frac{y}{x+y+z}\)

\(\frac{z}{z+x}>\frac{z}{x+y+z}\)

\(\Rightarrow A>\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}=\frac{x+y+z}{x+y+z}=1\)(1)

Lại có: \(\frac{x}{x+y}< \frac{x+y}{x+y+z}\)

\(\frac{y}{y+z}< \frac{y+z}{x+y+z}\)

\(\frac{z}{z+x}< \frac{z+x}{x+y+z}\)

\(\Rightarrow A< \frac{x+y}{x+y+z}+\frac{y+z}{x+y+z}+\frac{z+x}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)(2)

Từ (1) và (2) suy ra 1 < A < 2 

Vậy A không phải là số nguyên

TH1: \(x+y+z+t\ne0\) 

Áp dụng t/c dtsbn ta có:

\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}=\dfrac{x+y+z+t}{3\left(x+y+z+t\right)}=\dfrac{1}{3}\)\(\dfrac{x}{y+z+t}=\dfrac{1}{3}\Rightarrow3x=y+z+t\Rightarrow4x=x+y+z+t\\ \dfrac{y}{z+t+x}=\dfrac{1}{3}\Rightarrow3y=x+z+t\Rightarrow4y=x+y+z+t\\ \dfrac{z}{t+x+y}=\dfrac{1}{3}\Rightarrow3z=x+y+t\Rightarrow4z=x+y+z+t\\ \dfrac{t}{x+y+z}=\dfrac{1}{3}\Rightarrow3t=x+y+z\Rightarrow4t=x+y+z+t\)
\(\Rightarrow4x=4y=4z=4t\\ \Rightarrow x=y=z=t\)

\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\\ =1+1+1+1\\ =4\)

TH2: \(x+y+z+t=0\) 

\(\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(x+t\right)\\z+t=-\left(x+y\right)\\t+x=-\left(y+z\right)\end{matrix}\right.\)

\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\\ =\dfrac{-\left(z+t\right)}{z+t}+\dfrac{-\left(t+x\right)}{t+x}+\dfrac{-\left(x+y\right)}{x+y}+\dfrac{-\left(y+z\right)}{y+z}\\ =-1-1-1-1\\ =-4\)

Vì x, y, z là các số nguyên dương

Ta có: x/x+y>x/x+y+z

A = \(\frac{x+y-y}{x+y}+\frac{y+z-z}{y+z}+\frac{z+x-x}{x+z}\)

A=3 \(-\left(\frac{x}{x+z}+\frac{y}{x+y}+\frac{z}{y+z}\right)\)

mà \(\frac{x}{x+z}>\frac{x}{x+y+z};\frac{y}{y+z}>\frac{y}{x+y+z};\frac{z}{x+z}>\frac{z}{x+y+z}\)

=> A <2 (1)

mặt khác A=\(\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{x+z}\)

mà \(\frac{x}{x+y}>\frac{x}{x+y+z};\frac{y}{y+z}>\frac{y}{x+y+z};\frac{z}{x+z}>\frac{z}{x+y+z}\)

=> A >1 (2)

từ (1) và (2) => 1<A<2 => A ko phải là số nguyên

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