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Ta có: \(x+\frac{1}{y};y+\frac{1}{x}\) thuộc Z
=> \(\left(x+\frac{1}{y}\right)\left(y+\frac{1}{x}\right)=xy+x.\frac{1}{x}+\frac{1}{y}.y+\frac{1}{xy}=xy+\frac{1}{xy}=xy+\frac{1}{xy}\) thuộc Z
=> \(\left(xy+\frac{1}{xy}\right)^2=x^2y^2+2xy\frac{1}{xy}+\frac{1}{x^2y^2}=x^2y^2+\frac{1}{x^2y^2}+2\) thuộc Z
=> \(x^2y^2+\frac{1}{x^2y^2}\) thuộc Z
\(9x^2y^2+y^2-6xy-2y+2\)
\(=\left(9x^2y^2-6xy+1\right)+\left(y^2-2y+1\right)\)
\(=\left(3xy-1\right)^2+\left(y-1\right)^2\ge0\forall x,y\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}3xy-1=0\\y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy=\frac{1}{3}\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\y=1\end{matrix}\right.\)
Sửa lại đề : tính \(A=\frac{yz}{x^2+2yz}+\frac{xz}{y^2+2xz}+\frac{xy}{z^2+2xy}\)
Từ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Leftrightarrow\frac{xy+yz+xz}{xyz}=0\Rightarrow xy+yz+xz=0\)
\(\Rightarrow yz=-xy-xz\)
\(\Rightarrow x^2+2yz=x^2+yz-xy-xz=x\left(x-y\right)-z\left(x-y\right)=\left(x-z\right)\left(x-y\right)\)
CM tương tự ta cx có : \(\hept{\begin{cases}y^2+2xz=\left(y-x\right)\left(y-z\right)\\z^2+2xy=\left(z-x\right)\left(z-y\right)\end{cases}}\)
\(\Rightarrow A=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(y-x\right)\left(y-z\right)}+\frac{xy}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{yz\left(y-z\right)-xz\left(x-y-z+y\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{yz\left(y-z\right)+xz\left(z-y\right)-xz\left(x-y\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{\left(y-z\right)\left(yz-xz\right)+\left(x-y\right)\left(xy-xz\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{\left(y-z\right)\left(y-x\right)z+\left(x-y\right)\left(y-z\right)x}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(=\frac{\left(y-z\right)\left(x-y\right)\left(x-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
\(A=\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}\)
\(=\frac{x^4}{xy+2zx}+\frac{y^4}{yz+2xy}+\frac{z^4}{zx+2yz}\)
\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\frac{x^2+y^2+z^2}{3}=\frac{1}{3}\)
Xét \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
<=> \(a^2+b^2\ge2ab\) (luôn đúng)
Dấu bằng xảy ra khi a=b
Áp dụng ta có
\(\frac{1}{x+3y}+\frac{1}{y+2z+x}\ge\frac{4}{2\left(x+2y+z\right)}=\frac{2}{x+2y+z}\)
\(\frac{1}{y+3z}+\frac{1}{z+2x+y}\ge\frac{2}{x+y+2z}\)
\(\frac{1}{z+3x}+\frac{1}{x+2y+z}\ge\frac{2}{2x+y+z}\)
Cộng các vế của các bđt trên
=> ĐPCM
Dấu bằng xảy ra khi x=y=z