Từ x+y+z=3 ta có:

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\frac{\Leftrightarrow xy+yz+zx}{xyz}=\frac{1}{x+y+z}\)

Nhân chéo ta có:

\(\left(xy+yz+zx\right)\left(x+y+z\right)=xyz\)

\(\Leftrightarrow x^2y+xyz+x^2z+y^2x+y^2z+xyz+xyz+z^2y+z^2x=xyz\)

\(\Leftrightarrow x^2y+x^2z+y^2z+y^2x+z^2x+z^2y+2xyz=0\)

\(\Leftrightarrow\left(x^2y+x^2z+y^2x+xyz\right)+\left(y^2z+z^2x+z^2y+xyz\right)=0\)

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

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

\(\Leftrightarrow\left(x+z\right)\left[\left(xy+y^2\right)+\left(xz+yz\right)\right]=0\)

\(\Leftrightarrow\left(x+z\right)\left[y\left(x+y\right)+z\left(x+y\right)\right]=0\)

\(\Leftrightarrow\left(x+z\right)\left(y+z\right)\left(x+y\right)=0\)

Suy ra x+z=0 hoặc y+z=0 hoặc x+y=0

Với x+z=0 ta đc y=3

Với y+z=0 ta đc x=3

Với x+y=0 ta đc z=3

Từ đó suy ra đccm

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{x+y+z}-\frac{1}{z}\)

\(\Leftrightarrow\frac{x+y}{xy}=\frac{-x-y}{\left(x+y+z\right)z}\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{\left(x+y+z\right)z}\right)=0\)

\(+,x+y=0\Rightarrow x=-y\Rightarrow\text{đpcm}\)

\(+,\frac{1}{xy}+\frac{1}{\left(x+y+z\right)z}=0\Leftrightarrow\frac{xy+xz+yz+z^2}{xyz\left(x+y+z\right)}=0\Leftrightarrow\frac{x\left(y+z\right)+z\left(z+y\right)}{xyz\left(x+y+z\right)}=0\)

\(\Leftrightarrow\frac{\left(y+z\right)^2}{xyz\left(x+y+z\right)}=0\Rightarrow y+z=0\Rightarrow z=-y\Rightarrow\text{đpcm}\)

\(\text{Vậy ta có điều phải chứng minh }\)

     \(x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

\(\Leftrightarrow\)\(x+y+z=\frac{xy+yz+xz}{xyz}\)

\(\Leftrightarrow\)\(x+y+z=xy+yz+xz\)   (vì    xyz = 1 )

Ta có:      \(\left(xyz-1\right)+\left(x+y+z\right)-\left(xy+yz+xz\right)=0\)

\(\Leftrightarrow\)\(\left(xyz-xy\right)-\left(xz-x\right)-\left(yz-y\right)+\left(z-1\right)=0\)

\(\Leftrightarrow\)\(xy\left(z-1\right)-x\left(z-1\right)-y\left(z-1\right)+\left(z-1\right)=0\)

\(\Leftrightarrow\)\(\left(z-1\right)\left(x-1\right)\left(y-1\right)=0\)    (mk lm hơi tắt, thông cảm)

\(\Leftrightarrow\)  \(x-1=0\)            \(\Leftrightarrow\)      \(x=1\)

hoặc    \(y-1=0\)             \(\Leftrightarrow\)     \(y=1\)

hoặc    \(z-1=0\)             \(\Leftrightarrow\)     \(z=1\)

Vậy....

Áp dụng bđt Cauchy schwarz:

=> 1/x+1/y+4/z+16/t >= [(1+1+2+4)^2] / x+y+z+t=8^2/(x+y+z+t)=64/1=64

=> đpcm.

Áp dụng BĐT Svac - xơ:

\(\frac{1}{x}+\frac{1}{y}+\frac{4}{z}+\frac{16}{t}\ge\frac{\left(1+1+2+4\right)^2}{x+y+z+t}=\frac{64}{1}=64\)

(Dấu "="\(\Leftrightarrow x=y=\frac{1}{22};z=\frac{2}{11};t=\frac{8}{11}\))

*Áp dụng Cosi với x,y>0 ta có:

\(x+y\ge2\sqrt{xy}\left(1\right)\)

\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\left(2\right)\)

Nhân (1),(2) có: \(\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\RightarrowĐPCM\)

**\(\frac{1}{xy}+\frac{1}{x\left(x+y\right)}+\frac{1}{y\left(x+y\right)}+\frac{1}{x^2+y^2}\)

Ta có: \(\frac{1}{x\left(x+y\right)}+\frac{1}{y\left(x+y\right)}\ge\frac{4}{x^2+2xy+y^2}=4\)

Có: \(\frac{1}{x^2+xy}+\frac{1}{y^2+xy}\ge\frac{4}{\left(x+y\right)^2}\le4\)

Theo Cosi ta có: \(xy\le\left(\frac{x+y}{2}\right)^2\)

\(\Rightarrow\frac{1}{xy}\ge\left(\frac{2}{x+y}\right)^2\ge\left(\frac{2}{1}\right)^2=4\)

Áp dụng Cosi ta có: \(2xy\left(x^2+y^2\right)\le\left(\frac{x^2+2xy+y^2}{2}\right)^2=\frac{\left(x+y\right)^4}{4}\le\frac{1}{4}\)

\(\Rightarrow xy\left(x^2+y^2\right)\le\frac{1}{8}\)(1)

Mà ta có ở trên: \(xy\le\frac{\left(x+y\right)^2}{4}\le\frac{1}{4}\)(2)

Từ (1) và (2) ta có: \(x^2+y^2\le\frac{1}{2}\Rightarrow\frac{1}{x^2+y^2}\ge2\)

Vậy Ta có: \(\frac{1}{xy}+\frac{1}{x^2+xy}+\frac{1}{y^2+xy}+\frac{1}{x^2+y^2}\ge4+4+2=10\)

Với x=y=1/2