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

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 viết lại bất đẳng thức cần chứng minh thành: \(\frac{1}{\sqrt{xy}-4}+\frac{1}{\sqrt{yz}-4}+\frac{1}{\sqrt{zx}-4}\ge-1\)(*)

Theo BĐT Cauchy, ta có: \(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\le\frac{x+y}{2}+\frac{y+z}{2}+\frac{z+x}{2}=x+y+z\)

Mà ta có: \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=9\Rightarrow x+y+z\le3\)nên \(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\le3\)

Theo BĐT Bunyakovsky dạng phân thức: \(\frac{1}{\sqrt{xy}-4}+\frac{1}{\sqrt{yz}-4}+\frac{1}{\sqrt{zx}-4}\)\(\ge\frac{9}{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}-12}\ge\frac{9}{3-12}=-1\)

Suy ra (*) đúng

Vậy bất đẳng thức được chứng minh

Đẳng thức xảy ra khi x = y = z = 1

Ine CTV

dễ thấy \(x,y,z< \sqrt{3}\)\(\Rightarrow\)\(\sqrt{xy}-4< 0\); ... 

cauchy-schwarz chỉ dùng cho mẫu dương nha em, bài này lúc trước anh cũng lam sai, noi trước để đừng lục lại :D

ủa đây là toám lớp 1 hả anh

cauchy phần mẫu @@

\(\frac{x^2}{y+1}+\frac{y+1}{4}\ge x;\frac{y^2}{z+1}+\frac{z+1}{4}\ge y;\frac{z^2}{x+1}+\frac{x+1}{4}\ge z\)

\(\Rightarrow VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{4}.2=\frac{3}{2}\)

Cho a, b, c mà bắt chứng minh x, y, z nên ko chứng minh đc là đúng òi:)

\(VT-VP=\Sigma_{cyc}\frac{\left(x-y\right)^4}{4xy\left(x^2+y^2\right)}\ge0\)

a,b,c??? chỗ nào vậy bé ?? :)))

helloo

Ta có \(1+x^2=x^2+xy+yz+xz=\left(x+z\right)\left(x+y\right)\)

Khi đó BĐT <=>

 \(\frac{1}{\left(x+y\right)\left(x+z\right)}+\frac{1}{\left(y+z\right)\left(x+z\right)}+\frac{1}{\left(x+y\right)\left(y+z\right)}\ge\frac{2}{3}\left(\frac{x}{\sqrt{\left(x+z\right)\left(x+y\right)}}+...\right)\)

<=> \(\frac{x+y+z}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\frac{x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}}{\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}}\right)^3\)

<=>\(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge\frac{1}{3}\left(x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}\right)^3\)

<=> \(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\right)^3\)(1)

Xét \(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge\frac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)

<=> \(9\left[xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\right]\ge8\left(xy\left(x+y\right)+xz\left(x+z\right)+yz\left(y+z\right)+3xyz\right)\)

<=> \(xy\left(y+x\right)+yz\left(y+z\right)+xz\left(x+z\right)\ge6xyz\)

<=> \(x\left(y-z\right)^2+z\left(x-y\right)^2+y\left(x-z\right)^2\ge0\)luôn đúng

Khi đó (1) <=> 

\(\left(x+y+z\right).\frac{2\sqrt{2}}{3}.\sqrt{x+y+z}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+....\right)^3\) 

<=> \(\sqrt{2\left(x+y+z\right)}\ge\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\)

Áp dụng buniacopxki cho vế phải ta có 

\(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\le\sqrt{\left(x+y+z\right)\left(3-xy-yz-xz\right)}\)

                                                                                                       \(=\sqrt{2\left(x+y+z\right)}\)

=> BĐT được CM

Dấu bằng xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)

Ta có : \(\frac{x}{x^2-yz+2010}+\frac{y}{y^2-xz+2010}+\frac{z}{z^2-xy+2010}\)

\(=\frac{x^2}{x^3-xyz+2010x}+\frac{y^2}{y^3-xyz+2010y}+\frac{z^2}{z^3-xyz+2010z}\)

\(\ge\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2010\left(x+y+z\right)}=\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+3\left(xy+yz+xz\right)\left(x+y+z\right)}\)

\(=\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3+3xy^2+3x^2y+3x^2z+3xz^2+3y^2z+3yz^2}=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\frac{1}{x+y+z}\)

Ta có:

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

\(1+y^2=xy+yz+xz+y^2=\left(y+z\right)\left(x+y\right)\)

\(1+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\)

Thay vào A được:

\(P=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}+y\sqrt{\frac{\left(x+z\right)\left(y+z\right)\left(x+y\right)\left(x+z\right)}{\left(y+z\right)\left(x+y\right)}}\)\(+z\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(x+z\right)\left(y+z\right)}}\)

\(=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)

\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)

\(=xy+xz+xy+yz+xz+zy\)

\(=2\left(xy+yz+xz\right)\)

\(=2\)(do xy+yz+xz=1)

=>Đpcm

Dạng toán này rất nhiều bạn hỏi rồi: thay \(xy+yz+zx=1\) vào các căn thức rồi phân tích đa thức thành nhân tử.