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Ta có \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xyz}\left(x+y+z\right)=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{xyz}=4\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)(vì \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}>0\))
Mặt khác, ta có : \(\frac{1}{x+y+z}=2\) .
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
=> x+y = 0 hoặc y + z = 0 hoặc z + x = 0
Từ đó suy ra P = 0 (lí do vì x,y,z là các số mũ lẻ)
Có: \(x+y+z=\frac{1}{2}\Leftrightarrow2x+2y+2z=1\)
Mặt khác: \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{xyz}=4\)
\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2x+2y+2z}{xyz}=4\)
\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{zx}=4\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=4\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\) ( vì \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}>0\) )
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{\frac{1}{2}}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\frac{x+y}{xy}=\frac{1}{x+y+z}-\frac{1}{z}=\frac{-\left(x+y\right)}{z\left(x+y+z\right)}\)
\(\Leftrightarrow\left(x+y\right)\left(zx+yz+z^2\right)+xy\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(xy+yz+zx+z^2\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=0\\y+z=0\\z+x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x^{2021}+y^{2021}=0\\y^{2017}+z^{2017}=0\\z^{2019}+x^{2019}=0\end{matrix}\right.\)\(\Leftrightarrow Q=0\)
Vậy...
TA CÓ:
\(B=\frac{1}{\sqrt{x\left(y+2z\right)}}+\frac{1}{\sqrt{y\left(z+2x\right)}}+\frac{1}{\sqrt{z\left(x+2y\right)}}\ge\frac{1}{\frac{x+y+2z}{2}}+\frac{1}{\frac{y+z+2x}{2}}+\frac{1}{\frac{z+x+2y}{2}}\)
\(\ge\frac{\left(1+1+1\right)^2}{\frac{3}{2}\left(x+y+z\right)}=\frac{18}{3\sqrt{3}}=\frac{6}{\sqrt{3}}\)
DẤU BẰNG XẢY RA:\(\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\)
\(\frac{B}{\sqrt{3}}=\frac{1}{\sqrt{3x\left(y+2z\right)}}+\frac{1}{\sqrt{3y\left(z+2x\right)}}+\frac{1}{\sqrt{3z\left(x+2y\right)}}\)
\(\ge\frac{1}{\frac{3x+y+2z}{2}}+\frac{1}{\frac{3y+z+2x}{2}}+\frac{1}{\frac{3z+x+2y}{2}}\ge\frac{2\left(1+1+1\right)^2}{6\left(x+y+z\right)}=\frac{18}{6\sqrt{3}}\)
\(\Rightarrow B\ge\frac{18\sqrt{3}}{6\sqrt{3}}=3\)
Dấu "=" khi \(x=y=z=\frac{1}{\sqrt{3}}\)