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

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

\(x+y+z=xyz\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)

\(VT\le\dfrac{x}{2\sqrt{x^2yz}}+\dfrac{y}{2\sqrt{y^2zx}}+\dfrac{z}{2\sqrt{z^2xy}}\)

\(VT\le\dfrac{1}{2}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\le\dfrac{1}{2}\sqrt{3\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)}=\dfrac{\sqrt{3}}{2}\)

Dấu "=" xảy ra khi \(x=y=z=\sqrt{3}\)

\(VT=\dfrac{1}{z}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge\dfrac{1}{z}\left(\dfrac{4}{x+y}\right)=\dfrac{4}{z\left(x+y\right)}\ge\dfrac{16}{\left(z+x+y\right)^2}\ge16\) (đpcm)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\dfrac{1}{4};\dfrac{1}{4};\dfrac{1}{2}\right)\)

https://olm.vn/hoi-dap/detail/227981379332.html

Bạn tham khảo ở đây nhé.

http://diendantoanhoc.net/topic/160455-%C4%91%E1%BB%81-to%C3%A1n-v%C3%B2ng-2-tuy%E1%BB%83n-sinh-10-chuy%C3%AAn-b%C3%ACnh-thu%E1%BA%ADn-2016-2017/

bài của tui mà -_-