Pt tương đương:

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

Có: \(\sqrt{3\left(x^2+y^2+z^2\right)}\ge\sqrt{3\cdot3\left(xyz\right)^2}=3\)

Đồng thời:

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

Rồi cộng lại 

vì x²+y²+z²=1 mà x²+y²+z²>=xy+yz+xz suy ra 1>= xy+yz+xz

x²+y²+z²=1 suy ra (x-y)²=1-2xy-z² ,(y-z)²=1-2yz-x²,(x-z)²=(x-z)²=1-2xz-y²

\(\sqrt{3}+\frac{1}{2\sqrt{3}}[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2]=\)

\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)(do (x-y)²=1-2xy-z²(y-z)²=1-2yz-x²,(x-z)²=(x-z)²=1-2xz-y²)

theo bdt cosi ta có:

\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)

\(\le\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2z\sqrt{2xy}+2y\sqrt{2xz}+2x\sqrt{2yz}\right)]\)

\(\le\sqrt{3}+\frac{1}{2\sqrt{3}}[3-3\sqrt[3]{\left(2z\sqrt{2xy}.2y\sqrt{2xz}.2x\sqrt{2yz}\right)}\)

\(=\sqrt{3}+\frac{\sqrt{3}}{2}[1-2\sqrt{2}.\sqrt[3]{xyz^2}]\)\(=\sqrt{3}\left(1+\frac{1}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)=\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\)

suy ra 

\(\frac{x+y+z}{xy+yz+xz}\ge3.\sqrt[3]{xyz}\left(doxy+yz+xz\le1\right)\)

ta giả sử:

\(3\sqrt[3]{xyz}\ge\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\Leftrightarrow\sqrt{3}\ge\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\) mà \(\sqrt{3}>\frac{3}{2}\)

suy ra \(\frac{3}{2}\ge\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\)(luôn đúng) suy ra điều giả sử trên là đúng

hay \(3\sqrt[3]{xyz}\ge\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\)

mà \(\frac{x+y+z}{xy+yz+xz}\ge3.\sqrt[3]{xyz}\),\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)\(\le\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\)

suy ra \(\frac{x+y+z}{xy+yz+xz}\ge\)\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)

suy ra \(\frac{x+y+z}{xy+yz+xz}\ge\)\(\sqrt{3}+\frac{1}{2\sqrt{3}}[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2]\)(đpcm)

em mới có lớp 8, nếu em làm sai cho em xin lỗi nha anh

bạn ơi đk: 1 trong 3 số x,y,z là >=0 còn lại là >0 thì nó vẫn ra điều trên

BĐT <=> \(\sqrt{\frac{x+yz}{xyz}}+\sqrt{\frac{y+xz}{xyz}}+\sqrt{\frac{z+xy}{xyz}}\ge1+\sqrt{\frac{1}{xy}}+\sqrt{\frac{1}{yz}}+\sqrt{\frac{1}{xz}}\)

Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)

Khi đó \(a+b+c=1\)

BĐT <=>\(\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ab}\ge1+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)

Ta có \(\sqrt{a+bc}=\sqrt{a\left(a+b+c\right)+bc}=\sqrt{\left(a+b\right)\left(a+c\right)}\ge\sqrt{\left(a+\sqrt{bc}\right)^2}=a+\sqrt{bc}\)

Khi đó \(VT\ge a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=1+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=VP\)(ĐPCM)

Dấu bằng xảy ra khi x=y=z=3

BĐT cho tương đương với 

\(\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\ge1+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)

Với \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z};a+b+c=1\)

Ta có:

\(\sqrt{a+bc}=\sqrt{a\left(a+b+c\right)+bc}\)

\(=\sqrt{a^2+a\left(b+c\right)+bc}\ge\sqrt{a^2+2a\sqrt{bc}+bc}=a+\sqrt{bc}\)

Tương tự

\(\sqrt{b+ca}\ge b+\sqrt{ca};\sqrt{c+ab}\ge c+\sqrt{ab}\)

Từ đó ta có đpcm

Dấu "=" xảy ra khi x=y=z=3

rgghhjbjgtbjjjvjjjjjjgghhmbvkn  _{mm     ,,..,..guhhfxdgffdb ffd^hbvhgvg}ggg

gngn

\(\sqrt{x\left(1-y\right)\left(1-z\right)}=\sqrt{x\left(yz-y-z+1\right)}=\sqrt{x\left(yz-y-z+x+y+z+2\sqrt{xyz}\right)}\)

\(=\sqrt{x\left(yz+x+2\sqrt{xyz}\right)}=\sqrt{x^2+2x\sqrt{xyz}+xyz}=\sqrt{\left(x+\sqrt{xyz}\right)^2}\)

\(=x+\sqrt{xyz}\)

Tương tự: \(\sqrt{y\left(1-x\right)\left(1-z\right)}=y+\sqrt{xyz}\) ; \(\sqrt{z\left(1-x\right)\left(1-y\right)}=z+\sqrt{xyz}\)

\(\Rightarrow VT=x+y+z+3\sqrt{xyz}=1-2\sqrt{xyz}+3\sqrt{xyz}=1+\sqrt{xyz}\) (đpcm)

bạn làm đc ko

ko mới hỏi

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 \(x=y=z=\frac{1}{\sqrt{3}}\)

\(x^2+y^2+z^2\)

\(=\frac{x^2+y^2}{2}+\frac{y^2+z^2}{2}+\frac{z^2+x^2}{2}\)

\(\ge xy+yz+zx\)

\(=\frac{xy+yz}{2}+\frac{yz+zx}{2}+\frac{zx+xy}{2}\)

\(\ge\frac{2\sqrt{xy^2z}}{2}+\frac{2\sqrt{xyz^2}}{2}+\frac{2\sqrt{x^2yz}}{2}\)

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