Ta có \(x+y+z=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
\(\Leftrightarrow\left(x-2\sqrt{xy}+y\right)+\left(y-2\sqrt{yz}+z\right)+\left(z-2\sqrt{zx}+x\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{y}-\sqrt{z}\right)^2+\left(\sqrt{z}-\sqrt{x}\right)^2=0\) (*)
Với mọi \(x,y,z>0\) ta đều có
\(\left\{{}\begin{matrix}\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\\\left(\sqrt{y}-\sqrt{z}\right)^2\ge0\\\left(\sqrt{z}-\sqrt{x}\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{y}-\sqrt{z}\right)^2+\left(\sqrt{z}-\sqrt{x}\right)^2\ge0\)
Đẳng thức xảy ra tại (*) \(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x}-\sqrt{y}\right)^2=0\\\left(\sqrt{y}-\sqrt{z}\right)^2=0\\\left(\sqrt{z}-\sqrt{x}\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow x=y=z\)
\(x+y+z=\sqrt{xy}+\sqrt{xz}+\sqrt{yz}\)
\(\Leftrightarrow2x+2y+2z=2\sqrt{xy}+2\sqrt{xz}+2\sqrt{yz}\)
\(\Leftrightarrow2x+2y+2z-2\sqrt{xy}-2\sqrt{xz}-2\sqrt{yz}=0\)
\(\Leftrightarrow\left(x-2\sqrt{xy}+y\right)+\left(y-2\sqrt{yz}+z\right)+\left(z-2\sqrt{yz}+x\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{y}-\sqrt{z}\right)^2+\left(\sqrt{z}-\sqrt{x}\right)^2=0\)
Mà \(\left\{{}\begin{matrix}\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\\\left(\sqrt{y}-\sqrt{z}\right)^2\ge0\\\left(\sqrt{z}-\sqrt{x}\right)^2\ge0\end{matrix}\right.\Rightarrow\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{y}-\sqrt{z}\right)^2+\left(\sqrt{z}-\sqrt{x}\right)^2\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(\sqrt{x}-\sqrt{y}\right)^2=0\\\left(\sqrt{y}-\sqrt{z}\right)^2=0\\\left(\sqrt{z}-\sqrt{x}\right)^2=0\end{matrix}\right.\Rightarrow x=y=z\)
\(\Rightarrowđpcm\)
Giải:
Áp dụng BĐT AM - GM ta có:
\(x+y\ge2\sqrt{xy}\)
\(y+z\ge2\sqrt{yz}\)
\(x+z\ge2\sqrt{xz}\)
Cộng theo vế các BĐT trên ta có:
\(x+y+y+z+z+x\) \(\ge2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}\)
\(\Leftrightarrow2\left(x+y+z\right)\ge\) \(2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\right)\)
\(\Leftrightarrow x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)
