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Ta có: \(\left(x-1\right)^2\ge0\Leftrightarrow x^2+1\ge2x\)
Tương tự: \(y^2+1\ge2y;z^2+1\ge2z\)
\(x^2+y^2\ge2xy\) \(y^2+z^2\ge2yz\) \(z^2+x^2\ge2zx\)
\(\Rightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+zx\right)=12\)
\(\Leftrightarrow x^2+y^2+z^2\ge3\)
Dấu bằng xảy ra khi x=y=z=1
làm hơi tắt thông cảm
ủa,\(2\left(xy-yz+zx\right)\) mới đúng chứ nhể ?
\(x^2=\left(y+z\right)^2=y^2+2yz+z^2\Rightarrow2yz=x^2-y^2-z^2\)
\(x=y+z\Rightarrow x-y=z\Rightarrow x^2-2xy+y^2=z^2\Rightarrow x^2+y^2-z^2=2xy\)
\(x=y+z\Rightarrow y=x-z\Rightarrow y^2=x^2-2xz+z^2\Rightarrow x^2+z^2-y^2=2xz\)
Khi đó:
\(2xy-2yz+2zx=x^2+y^2-z^2-x^2+y^2+z^2+x^2+z^2-y^2=x^2+y^2+z^2\)
=> đpcm
Thêm một cách nhé!
\(x=y+z\)
=> \(y+z-x=0\)
=> \(\left(y+z-x\right)^2=0\)
=> \(\left(y+z\right)^2-2x\left(y+z\right)+x^2=0\)
=> \(x^2+y^2+z^2-2xy-2xz+2yz=0\)
=> \(2\left(xy-yz+xz\right)=x^2+y^2+z^2\)
Chứng minh \(x^2+y^2+z^2\ge xy+yz+xz\), Dấu "=" khi \(x=y=z\)
\(bdt\Leftrightarrow2x^2+2y^2+2z^2\ge2xy+2yz+2xz\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\forall x,y,z\in R\)
Dấu "=" khi \(\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{cases}\)\(\Leftrightarrow\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=y\\y=z\\z=x\end{cases}\)\(\Leftrightarrow x=y=z\)
Áp dụng vào bài ta có:
\(A=x^2+y^2+z^2\ge xy+yz+xz=12\)
Dấu "=' xảy ra khi \(\begin{cases}x=y=z\\xy+yz+xz=12\end{cases}\)\(\Leftrightarrow x=y=z=\pm2\)
Vậy \(Min_A=12\) khi \(x=y=z=\pm2\)
Vì xy + yz + zx = 1 ta có :
\(\frac{x-y}{z^2+1}+\frac{y-z}{x^2+1}+\frac{z-x}{y^2+1}=\frac{x-y}{z^2+xy+yz+zx}+\frac{y-z}{x^2+xy+yz+zx}+\frac{z-x}{y^2+xy+yz+zx}\)
\(=\frac{x-y}{\left(y+z\right)\left(z+x\right)}+\frac{y-z}{\left(x+y\right)\left(x+z\right)}+\frac{z-x}{\left(y+z\right)\left(x+y\right)}\)
\(=\frac{\left(x-y\right)\left(x+y\right)+\left(y-z\right)\left(y+z\right)+\left(x+z\right)\left(z-x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\frac{x^2-y^2+y^2-z^2+z^2-x^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{0}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=0\)(ĐPCM)
x2 + y2 + z2 = xy + yz + xz
<=> 2( x2 + y2 + z2 ) = 2( xy + yz + xz )
<=> 2x2 + 2y2 + 2z2 = 2xy + 2yz + 2xz
<=> 2x2 + 2y2 + 2z2 - 2xy - 2yz - 2xz = 0
<=> ( x2 - 2xy + y2 ) + ( y2 - 2yz + z2 ) + ( x2 - 2xz + z2 ) = 0
<=> ( x - y )2 + ( y - z )2 + ( x - z )2 = 0
Ta có : \(\hept{\begin{cases}\left(x-y\right)^2\\\left(y-z\right)^2\\\left(x-z\right)^2\end{cases}}\ge0\forall x,y,z\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\forall x,y,z\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y=0\\y-z=0\\x-z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=y\\y=z\\x=z\end{cases}}\Rightarrow x=y=z\left(đpcm\right)\)
Ta có: \(x^2+y^2+z^2=xy+yz+zx\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-z\right)^2=0\\\left(z-x\right)^2=0\end{cases}}\Rightarrow x=y=z\)
Ta có \(xy+xz+yz\le\frac{\left(x+y+z\right)^2}{3}\)
\(\Rightarrow x+y+z+\frac{\left(x+y+z\right)^2}{3}\ge6\)
\(\Rightarrow\left(x+y+z\right)^2+3\left(x+y+z\right)-18\ge0\)
\(\Rightarrow\left(x+y+z+6\right)\left(x+y+z-3\right)\ge0\)
\(\Rightarrow x+y+z-3\ge0\) (do \(x+y+z+6>0\))
\(\Rightarrow x+y+z\ge3\)
\(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\ge\frac{3^2}{3}=3\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
//Hoặc cách khác sử dụng AM-GM:
\(x^2+1\ge2x\) ; \(y^2+1\ge2y\); \(z^2+1\ge2z\);
\(x^2+y^2+z^2\ge xy+xz+yz\Rightarrow2x^2+2y^2+2z^2\ge2xy+2xz+2yz\)
Cộng vế với vế của 4 BĐT trên ta có:
\(3x^2+3y^2+3z^2+3\ge2\left(x+y+z+xy+xz+yz\right)=12\)
\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge9\)
\(\Rightarrow x^2+y^2+z^2\ge3\)
Dấu "=" xảy ra khi \(x=y=z=1\)