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\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\frac{xy+yz+zx}{xyz}=0\Leftrightarrow xy+yz+zx=0\)
\(\Leftrightarrow xy=-yz-zx;yz=-xy-zx;zx=-xy-yz\)
Ta có: x2+2yz=x2+yz+yz=x2+yz-xy-zx=x(x-y)-z(x-y)=(x-y)(x-z)
Tương tự: \(y^2+2xz=\left(y-x\right)\left(y-z\right);z^2+2xy=\left(z-x\right)\left(z-y\right)\)
A= \(\frac{yz}{x^2+2yz}+\frac{xz}{y^2+2xz}+\frac{xy}{z^2+2xy}\)=\(\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(y-x\right)\left(y-z\right)}+\frac{xy}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{yz\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{xz\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}+\frac{xy\left(x-y\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}\)
\(=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)\(=\frac{xy\left(x-y\right)-xz\left(x-y+y-z\right)+yz\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\frac{xy\left(x-y\right)-xz\left(x-y\right)-xz\left(y-z\right)+yz\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)\(=\frac{\left(xy-xz\right)\left(x-y\right)-\left(xz-yz\right)\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\frac{x\left(y-z\right)\left(x-y\right)-z\left(x-y\right)\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=\frac{\left(x-y\right)\left(y-z\right)\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}=1\)
Áp dụng bất đẳng thức Svacxo và bất đẳng thức \(\frac{1}{4ab}\ge\frac{1}{\left(a+b\right)^2}\)ta có :
\(Q=\frac{2}{x^2+y^2}+\frac{2}{2xy}+\frac{4}{2xy}=2\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{8}{4xy}\)
\(\ge2\frac{\left(1+1\right)^2}{\left(x+y\right)^2}+\frac{8}{\left(x+y\right)^2}=\frac{2.4}{2^2}+\frac{8}{2^2}=\frac{16}{4}=4\)
Dấu "=" xảy ra khi và chỉ khi \(x=y=1\)
Vậy min Q = 4 khi x = y = 1
