Lời giải:

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

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}\ge\frac{8^2}{4x+3y+z}\)

\(\Leftrightarrow\frac{4}{x}+\frac{3}{y}+\frac{1}{z}\ge\frac{64}{4x+3y+z}\)

Thiết lập tương tự với các phân thức còn lại:

\(\frac{4}{y}+\frac{3}{z}+\frac{1}{x}\ge\frac{64}{4y+3z+x}\)

\(\frac{4}{z}+\frac{3}{x}+\frac{1}{y}\ge\frac{64}{3x+y+4z}\)

Cộng theo vế: \(8\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge64\left(\frac{1}{4x+3y+z}+\frac{1}{x+4y+3z}+\frac{1}{3x+y+4z}\right)\)

\(\Leftrightarrow\frac{1}{4x+3y+z}+\frac{1}{x+4y+3z}+\frac{1}{3x+y+4z}\le\frac{1}{8}\)

Vậy GT:N của biểu thức là \(\frac{1}{8}\) khi \(x=y=z=3\)

Hay :D :) . Thanks chị 

Lời giải:

Từ $xy+yz+xz=xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1}{4x+3y+z}\leq \frac{1}{64}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}\right)\)

\(\frac{1}{x+4y+3z}\leq \frac{1}{64}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}+\frac{1}{z}\right)\)

\(\frac{1}{3x+y+4z}\leq \frac{1}{64}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}+\frac{1}{z}+\frac{1}{z}\right)\)

Cộng theo vế 3 BĐT trên và thu gọn ta được:

$A\leq \frac{1}{8}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{8}$

Vậy $A_{\max}=\frac{1}{8}$ khi $x=y=z=3$

\(xy+yz+xz=xyz\Rightarrow\)\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)

Áp dụng BĐT Cauchy Schwarz:

\(\dfrac{1}{4x+3y+z}\le\dfrac{1}{64}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

CMTT\(\Rightarrow\) \(M\le\dfrac{1}{64}\left(\dfrac{8}{x}+\dfrac{8}{y}+\dfrac{8}{z}\right)=\dfrac{1}{8}\)

Dấu''=" xảy ra\(\Leftrightarrow x=y=z=3\)

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

\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Tương tự:

\(\dfrac{y}{y+\sqrt{3y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\) ; \(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Cộng vế:

\(VT\le\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=1\)

\(P\ge\frac{x+y+z}{2}=\frac{\sqrt{\left(x+y+z\right)^2}}{2}\ge\frac{\sqrt{3\left(xy+yz+zx\right)}}{2}=\frac{\sqrt{3}}{2}\)

\("="\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\)

\(\sum\frac{x}{x+\sqrt{3x+yz}}=\sum\frac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}=\sum\frac{x}{x+\sqrt{\left(x+y\right)\left(x+z\right)}}\)

Sử dụng BĐT Cauchy-Schwarz, ta có

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

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