Ta có :

( x - 1 )²\(\ge\)0 => x² - 2x + 1 \(\ge\)0 => x² + 1 \(\ge\)2x

Tương tự ta có : y² + 1 \(\ge\)2y ; z² + 1 \(\ge\)2z

=> x² + y² + z² + 3 \(\ge\)2 ( x + y + z ) (1)

Lại có : ( x + y + z )² \(\ge\)0 => x² + y² + z² \(\ge\)2 ( xy + yz + zx ) (2)

Lấy (1) + (2) => 2 ( x² + y² + z² ) + 3 \(\ge\)2 ( x + y + z + xy + yz + zx )

<=> 2 ( x² + y² + z² ) \(\ge\)2.3033 - 3 = 6063

<=> x² + y² + z² \(\ge\)3031,5 > 2021 ( đpcm )

Bài 3 thì \(\le1\)

Bài 4 thì \(\ge\frac{3}{4}\) nhé

\(x^2+y^2+z^2=xy+yz+zx\\ \Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\\ \Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\Leftrightarrow x=y=z\\ \text{Mà }x+y+z=-3\Leftrightarrow x=y=z=-1\\ \Leftrightarrow B=1-1+1=1\)

Ta có : \(\left(x-y+z\right)^2=0\Leftrightarrow x^2+y^2+z^2-2xy+2xz-2yz=0\)

Mà \(x^2+y^2+z^2\ge0\) nên \(-2xy+2xz-2yz\le0\)

\(\Leftrightarrow-2\left(xy+yz-xz\right)\le0\)

\(\Rightarrow xy+yz-xz\ge0\)(đpcm)

Vì x-y+z=0   =>(x-y+z)²=0=>x²+y²+z²-2xy-2yz+2xz=0

=>x²+y²+z²=2xy+2yz-2xz mà x²+y²+z²\(\supseteq\)0

nên 2xy+2yz-2xz\(\supseteq\)0

=>xy+yz-xz\(\supseteq\)0

$P=\dfrac{x}{1-yz}+\dfrac{y}{1-xz}-\dfrac{z}{1+xy}$

Vì $z\le 0,\ x,y\ge 0$ nên $1-yz\le 1-z,\qquad 1-xz\le 1-z,\qquad 1+xy\le 1$

Suy ra $\dfrac{x}{1-yz}\ge \dfrac{x}{1-z}$

$\dfrac{y}{1-xz}\ge \dfrac{y}{1-z}$$-\dfrac{z}{1+xy}\ge -z$

Do đó $P\ge \dfrac{x+y}{1-z}-z$

$=\dfrac{x+y-z+z^2}{1-z}$$\ge \dfrac{\sqrt{x^2+y^2}-z+z^2}{1-z}$

Đặt $t=-z\ge0$.

Khi đó $x^2+y^2=1-t^2$ nên $P\ge \dfrac{\sqrt{1-t^2}+t+t^2}{1+t}$

Ta cần chứng minh $\sqrt{1-t^2}+t+t^2\ge 1+t$

$\Leftrightarrow \sqrt{1-t^2}\ge 1-t^2$

Điều này đúng vì $0\le 1-t^2\le 1$ và $\sqrt{u}\ge u\qquad (0\le u\le 1)$.

Suy ra $P\ge 1$.

Vậy $\dfrac{x}{1-yz}+\dfrac{y}{1-xz}-\dfrac{z}{1+xy}\ge 1.$