Đề sai:\(x+y+z=1\)
Đặt \(x^2+2xy=a;y^2+2xz=b;z^2+2xy=c\)
\(\Rightarrow a;b;c>0\) và \(a+b+c=\left(x+y+z\right)^2=1\)
\(\Rightarrow\frac{1}{x^2+2xy}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Áp dụng BĐT AM-GM ta có:\(a+b+c\ge3\sqrt[3]{abc}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge9\) vì \(a+b+c=1\)
\(\Rightarrow\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge9\left(đpcm\right)\)
Đề có j sai đâu đệ haizz
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{3}{\sqrt[3]{xyz}}\ge\frac{9}{x+y+z}\)
\(Apdung:\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge\frac{9}{\left(x+y+z\right)^2}\ge\frac{9}{1^2}=9\left(\text{đpcm}\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\frac{1}{x^2+2yz}=\frac{1}{y^2+2xz}=\frac{1}{z^2+2xy}\)
\(\Leftrightarrow x^2+2yz=y^2+2xz=z^2+2xy\)
\(\Leftrightarrow\hept{\begin{cases}x^2-y^2+2yz-2xz=0\\y^2-z^2+2xz-2xy=0\\z^2-x^2+2xy-2yz=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x-y\right)\left(x+y\right)-2z\left(x-y\right)=0\\\left(y-z\right)\left(z+y\right)-2x\left(y-z\right)=0\\\left(z-x\right)\left(z+x\right)-2y\left(z-x\right)=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x-y\right)^2-2z\left(x-y\right)=0\\\left(y-z\right)\left(z+y\right)-2x\left(y-z\right)=0\\\left(z-x\right)\left(z+x\right)-2y\left(z-x\right)=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x-y\right)\left(x+y\right)-2z\left(x-y\right)=0\\\left(y-z\right)\left(z+y\right)-2x\left(y-z\right)=0\\\left(z-x\right)\left(z+x\right)-2y\left(z-x\right)=0\end{cases}}\)
làm nốt
