Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}\\ =\dfrac{x+y+z+t}{y+z+t+z+t+x+t+x+y+x+y+z}\\ =\dfrac{x+y+z+t}{3x+3y+3z+3t}\\ =\dfrac{x+y+z+t}{3\left(x+y+z+t\right)}=\dfrac{1}{3}\)
Khi đó:
\(\dfrac{x}{y+z+t}=\dfrac{1}{3}\Rightarrow3x=y+z+t\left(1\right)\\ \dfrac{y}{z+t+x}=\dfrac{1}{3}\Rightarrow3y=z+t+x\left(2\right)\)
\(\dfrac{z}{t+x+y}=\dfrac{1}{3}\Rightarrow3z=t+x+y\left(3\right)\\ \dfrac{t}{x+y+z}=\dfrac{1}{3}\Rightarrow3t=x+y+z\left(4\right)\)
\(\left(1\right),\left(2\right),\left(3\right),\left(4\right)\Rightarrow3x=3y=3z=3t\\ \Rightarrow x=y=t=z\)
\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\)
\(\Rightarrow P=1+1+1+1=4\in Z\)
