\(\left|x-\frac{1}{2}\right|+\left|y+\frac{2}{3}\right|+\left|x^2+xz\right|=0\)

\(\Leftrightarrow\)\(\left|x-\frac{1}{2}\right|+\left|y+\frac{2}{3}\right|+\left|x\left(x+z\right)\right|=0\)

Ta có : 

\(\left|x-\frac{1}{2}\right|\ge0\)

\(\left|y+\frac{2}{3}\right|\ge0\)

\(\left|x\left(x+z\right)\right|\ge0\)

Mà \(\left|x-\frac{1}{2}\right|+\left|y+\frac{2}{3}\right|+\left|x\left(x+z\right)\right|=0\)

\(\Rightarrow\)\(\hept{\begin{cases}\left|x-\frac{1}{2}\right|=0\\\left|y+\frac{2}{3}\right|=0\\\left|x\left(x+z\right)\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{-2}{3}\\z=\frac{-1}{2}\end{cases}}}\)

Vậy \(x=\frac{1}{2}\)\(;\)\(y=\frac{-2}{3}\) và \(z=\frac{-1}{2}\)

Chúc bạn học tốt ~ 

Áp dụng BĐT Cô - si cho 3 bộ số không âm

\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(xz+1\right)^2}{x^2y^2z^2\left(yz+1\right)\left(xz+1\right)\left(xy+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)

Xét \(3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)

\(=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{xz+1}{z}\right)}\)

\(=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)

Áp dụng BĐT Cô - si

\(\Rightarrow\left\{\begin{matrix}y+\frac{1}{x}\ge2\sqrt{\frac{y}{x}}\\z+\frac{1}{y}\ge2\sqrt{\frac{z}{y}}\\x+\frac{1}{z}\ge2\sqrt{\frac{x}{z}}\end{matrix}\right.\)

\(\Rightarrow\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)\ge8\)

\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge3\sqrt[3]{8}\)

\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge6\)

\(\Leftrightarrow3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\ge6\)

Mà \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)

\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge6\)

Vậy GTNN của \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}=6\)

Bài này cô giáo giảng rùi mà

quên rồi

X= 1/2. Y=-1/4. X=1

Xin lỗi z=-1/2

Ta có \(P=\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}+\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}\)

\(=\frac{\frac{\left(yz+1\right)^2}{z^2}}{\frac{zx+1}{x}}+\frac{\frac{\left(zx+1\right)^2}{x^2}}{\frac{xy+1}{y}}+\frac{\frac{\left(xy+1\right)^2}{y^2}}{\frac{yz+1}{z}}\)

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

Áp dụng BĐT \(\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\frac{a_3^2}{b_3}\ge\frac{\left(a_1+a_2+a_3\right)^2}{b_1+b_2+b_3}\)

Dấu "=" xảy ra khi \(\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{c_3}\)

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

\(P\ge a+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

Áp dụng BĐT: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)

=> \(P\ge x+y+z+\frac{9}{x+y+z}=\left[x+y+z+\frac{9}{4\left(x+y+z\right)}\right]+\frac{27}{4\left(x+y+z\right)}\)

Ta có: \(x+y+z+\frac{9}{4\left(x+y+z\right)}\ge2\sqrt{\frac{9}{4}}=3;\frac{27}{4\left(x+y+z\right)}=\frac{27}{4\cdot\frac{3}{2}}=\frac{9}{2}\)

=> \(P\ge3+\frac{9}{2}=\frac{15}{2}\).

Dấu "=" xảy ra <=> x=y=z=\(\frac{1}{2}\)

Vậy Min_P=\(\frac{15}{2}\)đạt được khi x=y=z=\(\frac{1}{2}\)

Ta có:

\(P=\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}+\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}\)

\(=\frac{\left(\frac{yz+1}{z}\right)^2}{\left(\frac{zx+1}{x}\right)}+\frac{\left(\frac{zx+1}{x}\right)^2}{\left(\frac{xy+1}{y}\right)}+\frac{\left(\frac{xy+1}{y}\right)^2}{\left(\frac{yz+1}{z}\right)}\)

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

Áp dụng BĐT Bunhiacopxki dạng phân thức, ta có:

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

\(\ge\left(x+y+z\right)+\frac{9}{x+y+z}=\left(x+y+z\right)+\frac{9}{4\left(x+y+z\right)}\)

\(+\frac{27}{4\left(x+y+z\right)}\ge2\sqrt{\left(x+y+z\right).\frac{9}{4\left(x+y+z\right)}}+\frac{27}{4.\frac{3}{2}}=\frac{15}{2}\)(Áp dụng BĐT Cô - si cho 2 số không âm)

Đẳng thức xảy ra khi \(x=y=z=\frac{1}{2}\)

Một cách giải khác ( cách này em làm rùi giờ làm lại ạ ) cô Chi check em ạ :)

Áp dụng BĐT AM-GM ta có:

\(P=\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}+\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}\)

\(\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(zx+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}\)

\(=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}\)

\(=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)

Áp dụng tiếp BĐT AM-GM ta có:

\(y+\frac{1}{x}=y+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}\ge5\sqrt[5]{\frac{y}{256x^4}}\)

Tương tự \(z+\frac{1}{y}\ge5\sqrt[5]{\frac{z}{256y^4}};x+\frac{1}{z}\ge5\sqrt[5]{\frac{x}{256z^4}}\)

Sử dụng liên hoàn BĐT AM-GM ta có tiếp

\(P\ge3\sqrt[3]{5\sqrt[5]{\frac{y}{256x^4}}\cdot5\sqrt[5]{\frac{z}{256y^4}}\cdot5\sqrt[5]{\frac{x}{256z^4}}}\)

\(=3\sqrt[3]{125\sqrt[5]{\frac{xyz}{256^3\left(xyz\right)^4}}}\)

\(=15\sqrt[3]{\sqrt[5]{\frac{1}{256^3\left(xyz\right)^3}}}\)

\(=15\sqrt[15]{\frac{1}{256^3\left(xyz\right)^3}}\)

\(\ge15\sqrt[15]{\frac{1}{256^3\left(\frac{x+y+z}{3}\right)^9}}\)

\(\ge15\sqrt[15]{256^3\cdot\frac{1}{2^9}}=\frac{15}{2}\)

Dấu "='" xảy ra tại x=y=z=¹/₂