\(\frac{y+z+1+x+z+1+x+y-3}{x+y+z}\)=\(\frac{2\left(X+Y+Z\right)}{x+y+z}\)=2  =>x+y+z=\(\frac{1}{2}\)   tu lam di nhe

Áp dụng BĐT AM-GM ta có:

\(\hept{\begin{cases}\sqrt{xy}\le\frac{x+y}{2}\\\sqrt{yz}\le\frac{y+z}{2}\\\sqrt{xz}\le\frac{x+z}{2}\end{cases}}\). Cộng theo vế ta có:

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

Do đó ta có: \(x+y+z\ge1\).Áp dụng BĐT Cauchy-Schwarz dạng Engel ta cũng có:

\(A\ge\frac{\left(x+y+z\right)^2}{x+y+y+z+x+z}=\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{1}{2}\)

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

Hình như sai đề rồi bạn :

Có phải như thế này không :

\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+y}\)

Ta có\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}\)

\(=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}\)

\(=\dfrac{2x+2y+2z+1+2-3}{x+y+z}\)

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

Nên \(\dfrac{1}{x+y+z}=2\Rightarrow x+y+z=\dfrac{1}{2}\)

Ta lại có:

\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=2\)

\(\Leftrightarrow\dfrac{\left(x+y+z\right)-z+1}{x}=\dfrac{\left(x+y+z\right)-y+2}{y}=\dfrac{\left(x+y+z\right)-z-3}{z}=2\)

\(\Rightarrow\dfrac{\dfrac{1}{2}-x+1}{x}=\dfrac{\dfrac{1}{2}-y+2}{y}=\dfrac{\dfrac{1}{2}-z-3}{z}=2\)

\(\Rightarrow\dfrac{\dfrac{3}{2}-x}{x}=\dfrac{\dfrac{5}{2}-y}{y}=\dfrac{-z-\dfrac{5}{2}}{z}=2\)

\(\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{\dfrac{3}{2}-x}{x}\\\dfrac{\dfrac{5}{2}-y}{y}\\\dfrac{-z-\dfrac{5}{2}}{z}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=\dfrac{3}{2}-x\\2y=\dfrac{5}{2}-y\\2z=-z-\dfrac{5}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{6}\\z=\dfrac{5}{2}\end{matrix}\right.\)