e cunho tui ko ba

Ta có:\(\sqrt{x^2-xy+y^2}=\sqrt{\frac{1}{4}\left(x+y\right)^2+\frac{5}{4}\left(x-y\right)^2}\ge\frac{1}{2}\left(x+y\right)\)

Ttự,có:\(\sqrt{y^2-yz+z^2}\ge\frac{1}{2}\left(y+z\right);\sqrt{z^2-xz+x^2}\ge\frac{1}{2}\left(x+z\right)\)

\(\Rightarrow2S\ge\frac{x+y}{x+y+2z}+\frac{y+z}{y+z+2x}+\frac{z+x}{z+x+2y}\)

Đặt \(a=x+y+2z;b=y+z+2x;c=z+x+2y\)

Có:\(b+c-a=2\left(x+y\right);a+c-b=2\left(y+z\right);a+b-c=2\left(x+z\right)\)

\(\Rightarrow4S\ge\frac{b+c-a}{a}+\frac{a+c-b}{b}+\frac{a+b-c}{c}\)

\(=\frac{b}{a}+\frac{c}{a}+\frac{a}{b}+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}+1\)

\(\ge2+2+2+1=7\)

\(\Rightarrow S\ge\frac{7}{4}\)

\(S_{min}=\frac{7}{4}\Leftrightarrow x+y+2z=y+z+2x=x+z+2y\)\(\Leftrightarrow x=y=z\)

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

\(A=\sqrt{\frac{x^2}{2xyz.yz+xz.xy}}+\sqrt{\frac{y^2}{2xyz.xz+xy.yz}}+\sqrt{\frac{z^2}{2xyz.xy+xz.yz}}\)

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

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

Áp dụng bđt \(\sqrt{ab}\le\frac{a+b}{2}\) ta có:

\(2A\le\frac{x}{yz+xy}+\frac{x}{yz+xz}+\frac{y}{xz+xy}+\frac{y}{xz+yz}+\frac{z}{xy+yz}+\frac{z}{xy+xz}\)

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

Mà: \(xy+yz+xz=2xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)

\(\Rightarrow2A\le2\Rightarrow A\le1."="\Leftrightarrow a=b=c=\frac{3}{2}\)

KON 'NICHIWA ON" NANOKO: chào cô