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\(\dfrac{xy}{x+y}=\dfrac{yz}{y+z}=\dfrac{zx}{z+x}\\ \Rightarrow\dfrac{x+y}{xy}=\dfrac{y+z}{yz}=\dfrac{z+x}{zx}\\ \Rightarrow\dfrac{1}{y}+\dfrac{1}{x}=\dfrac{1}{z}+\dfrac{1}{y}=\dfrac{1}{x}+\dfrac{1}{z}\\ \Rightarrow\dfrac{1}{x}=\dfrac{1}{y}=\dfrac{1}{z}\\ \Rightarrow x=y=z\)
\(\Rightarrow P=\dfrac{xy+yz+zx}{x^2+y^2+z^2}=\dfrac{x^2+x^2+x^2}{x^2+x^2+x^2}=1\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Leftrightarrow\dfrac{yz+xz+xy}{xyz}=0\)
\(\Leftrightarrow xy+xz+yz=0\)
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\(x^2+2yz\)
\(=x^2+yz-xy-xz\)
\(=\left(x-y\right)\left(x-z\right)\)
Tương tự, ta có: \(y^2+2xz=\left(y-x\right)\left(y-z\right)\) và \(z^2+2xy=\left(z-x\right)\left(z-y\right)\)
\(A=\dfrac{yz}{\left(x-z\right)\left(x-y\right)}+\dfrac{xz}{\left(y-x\right)\left(y-z\right)}+\dfrac{xy}{\left(z-x\right)\left(z-y\right)}\)
\(A=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-z\right)\left(x-y\right)\left(y-z\right)}\)
\(=\dfrac{\left(x-z\right)\left(x-y\right)\left(y-z\right)}{\left(x-z\right)\left(x-y\right)\left(y-z\right)}\)
= 1
\(\dfrac{xy}{ay+bx}=\dfrac{yz}{bz+cy}=\dfrac{zx}{cx+az}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(1\right)\)
Ta có: \(\dfrac{xy}{ay+bx}=\dfrac{yz}{bz+cy}=\dfrac{zx}{cx+az}\)
\(\Rightarrow\dfrac{xyz}{ayz+bxz}=\dfrac{xyz}{bxz+cxy}=\dfrac{xyz}{cxy+ayz}\)
\(\Rightarrow ayz+bxz=bxz+cxy=cxy+ayz\)
\(\Rightarrow\left\{{}\begin{matrix}ayz+bxz=bxz+cxy\\ayz+bxz=cxy+ayz\\bxz+cxy=cxy+ayz\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}ayz=cxy\\bxz=cxy\\bxz=ayz\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}az=cx\\bz=cy\\bx=ay\end{matrix}\right.\left(2\right)\)
Thay (2) vào (1) ta có :
\(\dfrac{xy}{2ay}=\dfrac{yz}{2bz}=\dfrac{xz}{2cx}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\)
\(\Rightarrow\dfrac{x}{2a}=\dfrac{y}{2b}=\dfrac{z}{2c}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(3\right)\)
\(\Rightarrow\dfrac{x^2}{4a^2}=\dfrac{y^2}{4b^2}=\dfrac{z^2}{4c^2}=\dfrac{\left(x^2+y^2+z^2\right)^2}{\left(a^2+b^2+c^2\right)^2}=\)\(\dfrac{x^2+y^2+z^2}{4a^2+4b^2+4c^2}\)
\(\Rightarrow\dfrac{x^2+y^2+y^2}{a^2+b^2+c^2}=\dfrac{1}{4}\left(4\right).\)Thay (3) vào (2) ta có :
\(\dfrac{x}{2a}=\dfrac{y}{2b}=\dfrac{z}{2c}=\dfrac{1}{4}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{a}{2}\\y=\dfrac{b}{2}\\z=\dfrac{c}{2}\end{matrix}\right.\)
