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\(\Leftrightarrow\sqrt{4x^2+4xy+8y^2}+\sqrt{4y^2+4yz+8z^2}+\sqrt{4z^2+4zx+8x^2}\ge4\left(x+y+z\right)\)
Ta có:
\(VT=\sqrt{\left(2x+y\right)^2+\left(\sqrt{7}y\right)^2}+\sqrt{\left(2y+z\right)^2+\left(\sqrt{7}z\right)^2}+\sqrt{\left(2z+x\right)^2+\left(\sqrt{7}x\right)^2}\)
\(VT\ge\sqrt{\left(2x+y+2y+z+2z+x\right)^2+\left(\sqrt{7}x+\sqrt{7}y+\sqrt{7}z\right)^2}\)
\(VT\ge\sqrt{16\left(x+y+z\right)^2}=4\left(x+y+z\right)\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z\)
BĐT Mincopxki:
\(\sqrt{x^2+a^2}+\sqrt{y^2+b^2}+\sqrt{z^2+c^2}\ge\sqrt{\left(x+y+z\right)^2+\left(a+b+c\right)^2}\)
ta có:
(x+y+z)2=x2+y2+z2+2xy+2xz+2yz
<=>(x+y+z)2=x2+y2+z2+2.(xy+xz+yz)
thay x+y+z=0 và xy+xz+yz=0 ta được:
02=x2+y2+z2=2.0
<=>x2+y2+z2=0
mà x2;y2;z2\(\ge\)0 nên
=>x=y=z=0 thì x2+y2+z2=0
vậy với x+y++z=0 và xy+yz+zx=0 thì x=y=z
Ta có :
\(\left(x+y+z\right)^2\)
\(=x^2+y^2+z^2+2\left(xy+yz+zx\right)\)
\(\Rightarrow0=x^2+y^2+z^2+2.0\)
\(\Rightarrow0=x^2+y^2+z^2\)
Vậy \(x=y=z\left(=0\right)\)(đpcm)
\(\dfrac{x-y}{z^2+1}=\dfrac{x-y}{z^2+xy+yz+zx}=\dfrac{x-y}{z\left(z+y\right)+x\left(z+y\right)}=\dfrac{x-y}{\left(x+z\right)\left(z+y\right)}\)
Tương tự: \(\dfrac{y-z}{x^2+1}=\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}\);\(\dfrac{z-x}{y^2+1}=\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)
Cộng vế với vế \(\Rightarrow VT=\dfrac{x-y}{\left(x+z\right)\left(y+z\right)}+\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}+\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)+\left(y-z\right)\left(y+z\right)+\left(z-x\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\dfrac{x^2-y^2+y^2-z^2+z^2-x^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=0\)(đpcm)
\(\left\{{}\begin{matrix}x+y+z=0\\xy+yz+zx=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+y+z\right)^2=0\\2\left(xy+yz+zx\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x^2+y^2+z^2+2xy+2yz+2zx=0\\2xy+2yz+2zx=0\end{matrix}\right.\)
\(\Rightarrow x^2+y^2+z^2+2xy+2yz+2zx-2xy-2yz-2zx=0\)
\(\Rightarrow x^2+y^2+z^2=0\)
\(\left\{{}\begin{matrix}x^2\ge0\forall x\\y^2\ge0\forall y\\z^2\ge0\forall z\end{matrix}\right.\)
Nên: \(x^2+y^2+z^2\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}x^2=0\\y^2=0\\z^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\\z=0\end{matrix}\right.\)
Vậy \(x=y=z=0\)
Ta có điều phải chứng minh
ta có x+y+z=0 suy ra (x+y+z)2=0
do đó x2+y2+z2=0(vì xy+yz+xz=0)
vì thế x=y=z