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

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)

1111111111111111111

\(VT=\Sigma\frac{xy+yz+zx}{xy}=3+\Sigma\frac{z\left(x+y\right)}{xy}\)

Đến đây để ý \(\frac{1}{2}\left[\frac{z\left(x+y\right)}{xy}+\frac{y\left(z+x\right)}{zx}\right]\ge\sqrt{\frac{\left(z+x\right)\left(x+y\right)}{x^2}}\left(\text{AM - GM}\right)\)

Là xong.

ủa đây là toám lớp 1 hả anh

cauchy phần mẫu @@

Forever_Alone tên là Anh nhưng ko bt họ

Đặt vế trái là P

\(P=\frac{x^2y^2+y^2z^2+z^2x^2}{xyz}\Rightarrow P^2=\frac{x^4y^4+y^4z^4+z^4x^4+2x^2y^2z^2\left(x^2+y^2+z^2\right)}{x^2y^2z^2}\)

\(P^2\ge\frac{x^2y^2z^2\left(x^2+y^2+z^2\right)+6x^2y^2z^2}{x^2y^2z^2}=\frac{9x^2y^2z^2}{x^2y^2z^2}=9\)

\(\Rightarrow P\ge3\)

Dấu "=" xảy ra khi \(x=y=1=z\)

x+y+z=3 ms lm đc

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) 

ta caàn chứng minh bđt 

\(\frac{x}{x+yz}+\frac{y}{y+zx}\ge\frac{x}{x+xz}+\frac{y}{y+yz}=\frac{1}{1+z}+\frac{1}{1+z}=\frac{2}{1+z}\)

tương tự + vào, dùng svác sơ

Áp dụng BĐT Cô-si dạng Engel,ta có :

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

Mà \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le x+y+z\)

\(\Rightarrow\)\(\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\ge\frac{3}{2}\)

Dấu "=" xảy ra khi x = y = z = \(\frac{3}{2}\)

nhầm sửa x = y = z = 1 nha

Áp dụng bất đẳng thức cộng mẫu số 

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

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

Áp dụng bất đẳng thức Cacuchy cho 2 bộ số thực không âm

\(\Rightarrow\hept{\begin{cases}\sqrt{yz}\le\frac{y+z}{2}\\\sqrt{xz}\le\frac{x+z}{2}\\\sqrt{xy}\le\frac{x+y}{2}\end{cases}}\)

\(\Rightarrow\sqrt{yz}+\sqrt{xz}+\sqrt{xy}\le\frac{y+z}{2}+\frac{x+z}{2}+\frac{x+y}{2}\)

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

\(\Rightarrow\sqrt{yz}+\sqrt{xz}+\sqrt{xy}\le x+y+z\)

\(\Rightarrow x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le2\left(x+y+z\right)\)

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

Ta có : \(x+y+z\ge3\)

\(\Rightarrow\frac{x+y+z}{2}\ge\frac{3}{2}\)

\(\Rightarrow\frac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{xz}+\sqrt{yz}}\ge\frac{3}{2}\)

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

\(\Rightarrow\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{xz}}+\frac{z^2}{z+\sqrt{xy}}\ge\frac{3}{2}\left(đpcm\right)\)

Chúc bạn học tốt !!!