Ta có : 

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

\(=x^2+xz+xy+yz+y^2+xy+zy+xz\)

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

\(2\left(x+z\right)\left(z+y\right)\)

\(=2\left(xz+z^2+xy+zy\right)\)

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

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

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

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

Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Leftrightarrow\frac{xy+yz+zx}{xyz}=0\)

\(\Rightarrow xy+yz+zx=0\)

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

\(\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2\)

Vậy \(\left(x+y+z\right)^2=x^2+y^2+z^2\)

ra giong minh phet

Lời giải:

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

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

\(\Leftrightarrow \frac{x^2}{y+z}+\frac{xy}{x+z}+\frac{xz}{x+y}+\frac{xy}{y+z}+\frac{y^2}{x+z}+\frac{zy}{x+y}+\frac{xz}{y+z}+\frac{zy}{x+z}+\frac{z^2}{x+y}=2(x+y+z)\)

\(\Leftrightarrow \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}+\frac{xy+zy}{x+z}+\frac{xz+yz}{x+y}+\frac{xy+xz}{y+z}=2(x+y+z)\)

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

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

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

+) TH1: x + y + z = 0 => x + y = -z ; x + z = -y; y + z = -x

Do đó: \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=\frac{x}{-x}+\frac{y}{-y}=\frac{z}{-z}=-3\)\(\ne1\)loại

+) TH2: x + y + z \(\ne0\)

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

<=> \(\frac{x\left(x+y+z\right)}{y+z}+\frac{y\left(x+y+z\right)}{z+x}+\frac{z\left(x+y+z\right)}{x+y}=x+y+z\)

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

<=> \(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=0\)( đpcm)

Áp dụng BĐT Cô-si cho 2 số dương, ta có:

\(18x+\frac{2}{x}\ge2\sqrt{18x.\frac{2}{x}}=12\)

Chứng minh tương tự, ta có

\(18y+\frac{2}{y}\ge12\)

\(18z+\frac{2}{z}\ge12\)

Từ đó suy ra \(18\left(x+y+z\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge36\)(*)

Lại có \(x+y+z\le1\Rightarrow-\left(x+y+z\right)\ge-1\)(**)

Từ (*) và (**) suy ra \(18\left(x+y+z\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-\left(x+y+z\right)\ge36-1\)

                           \(\Leftrightarrow17\left(x+y+z\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge35\)

Vậy \(17\left(x+y+z\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge35\)với \(x+y+z\le1\)