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)

`x+y+z>=0` là chưa đủ phải là `x,y,z>=0` mới đúng.

`x+y+z>=sqrt{xy}+sqrt{yz}+sqrt{zx}`

`<=>2x+2y+2z>=2sqrt{xy}+2sqrt{yz}+2sqrt{zx}`

`<=>x-2sqrt{xy}+y+y-2sqrt{yz}+z+z-2sqrt{zx}+x>=0`

`<=>(sqrtx-sqrty)^2+(sqrty-sqrtz)^2+(sqrtz-sqrtx)^2>=0` luôn đúng

Dấu `"="<=>x=y=z`

Áp dụng bdt Co-si, ta có:

\(x+y\ge2\sqrt{xy}\)

\(y+z\ge2\sqrt{yz}\)

\(z+x\ge2\sqrt{xz}\)

=> 2(x+y+z) \(\ge2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)

=> đpcm

Áp dụng BĐT AM-GM ta có:

\(\frac{\left(y+z\right)\sqrt{yz}}{x}\ge\frac{2\sqrt{yz}\cdot\sqrt{yz}}{x}=\frac{2\sqrt{\left(yz\right)^2}}{x}=\frac{2yz}{x}\)

Tương tự cho 2 BĐT còn lại ta cũng có

\(\frac{\left(x+y\right)\sqrt{xy}}{z}\ge\frac{2xy}{z};\frac{\left(x+z\right)\sqrt{xz}}{y}\ge\frac{2xz}{y}\)

\(\Leftrightarrow\frac{\left(y+z\right)\sqrt{yz}}{x}+\frac{\left(x+y\right)\sqrt{xy}}{z}+\frac{\left(x+z\right)\sqrt{xz}}{y}\ge\frac{2xy}{z}+\frac{2yz}{x}+\frac{2xz}{y}\)

Cần chứng minh \(\frac{2xy}{z}+\frac{2yz}{x}+\frac{2xz}{y}\ge2\left(x+y+z\right)\)

\(\Leftrightarrow\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\ge x+y+z\)

Áp dụng BĐT AM-GM:

\(\frac{xy}{z}+\frac{yz}{x}\ge2\sqrt{\frac{xy}{z}\cdot\frac{yz}{x}}=2\sqrt{y^2}=2y\)

Tương tự rồi cộng theo vế ta có ĐPCM

Khi \(x=y=z\)

\(\left\{{}\begin{matrix}x+y+z=0\\xy+yz+zx=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x+y+z\right)^2=0\\2\left(xy+yz+zx\right)=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2+y^2+z^2+2xy+2yz+2zx=0\\2xy+2yz+2zx=0\end{matrix}\right.\)

\(\Rightarrow x^2+y^2+z^2+2xy+2yz+2zx-2xy-2yz-2zx=0\)

\(\Rightarrow x^2+y^2+z^2=0\)

\(\left\{{}\begin{matrix}x^2\ge0\forall x\\y^2\ge0\forall y\\z^2\ge0\forall z\end{matrix}\right.\)

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

\(\dfrac{x-y}{z^2+1}=\dfrac{x-y}{z^2+xy+yz+zx}=\dfrac{x-y}{z\left(z+y\right)+x\left(z+y\right)}=\dfrac{x-y}{\left(x+z\right)\left(z+y\right)}\)

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)

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