Ta có
\(x-y=\left(by+cz\right)-\left(ax+cz\right)=by-ax\)
\(\Leftrightarrow x\cdot\left(a+1\right)=y\cdot\left(b+1\right)\)
\(y-z=\left(ax+cz\right)-\left(ax+by\right)=cz-by\)
\(\Leftrightarrow z\cdot\left(c+1\right)=y\cdot\left(b+1\right)\)
\(x-z=\left(by+cz\right)-\left(ax+by\right)=cz-ax\)
\(\Leftrightarrow x\cdot\left(a+1\right)=z\cdot\left(c+1\right)\)
\(\Rightarrow x\cdot\left(a+1\right)=z\cdot\left(c+1\right)=y\left(b+1\right)\)
Đặt \(x\cdot\left(a+1\right)=z\cdot\left(c+1\right)=y\left(b+1\right)=k\)
\(\Rightarrow\left\{{}\begin{matrix}a+1=\dfrac{k}{x}\\b+1=\dfrac{k}{y}\\c+1=\dfrac{k}{z}\end{matrix}\right.\)
Thay vào A, ta có :
\(A=\dfrac{1}{\dfrac{k}{x}}+\dfrac{1}{\dfrac{k}{y}}+\dfrac{1}{\dfrac{k}{z}}\)
\(=\dfrac{x}{k}+\dfrac{y}{k}+\dfrac{z}{k}\)
=\(\dfrac{x+y+z}{k}\)
Vì z = ax + by; x = cz + by; y = ax + cz nen :
\(k=z\cdot\left(c+1\right)=cz+z=cz+ax+by\)
\(\Rightarrow A=\dfrac{2\cdot\left(ax+by+czz\right)}{ax+by+cz}=2\)
⇒ĐPCM