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

Áp dụng bất đẳng thức AM - GM:

\(P\ge3\sqrt[3]{\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}\).

Áp dụng bất đẳng thức AM - GM ta có:

\(xy+1=xy+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\ge5\sqrt[5]{\dfrac{xy}{4^4}}\).

Tương tự: \(yz+1\ge5\sqrt[5]{\dfrac{yz}{4^4}};zx+1\ge5\sqrt[5]{\dfrac{zx}{4^4}}\).

Do đó \(\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)\ge125\sqrt[5]{\dfrac{\left(xyz\right)^2}{4^{12}}}\)

\(\Rightarrow\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{1}{4^{12}\left(xyz\right)^3}}\).

Mà \(xyz\le\dfrac{\left(x+y+z\right)^3}{27}=\dfrac{1}{8}\)

Nên \(\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{8^3}{4^{12}}}=125\sqrt[5]{\dfrac{1}{2^{15}}}=\dfrac{125}{8}\)

\(\Rightarrow P\ge\dfrac{15}{2}\).

Vậy...

Áp dụng bất đẳng thức AM - GM:

P≥33√(xy+1)(yz+1)(zx+1)xyz.

Áp dụng bất đẳng thức AM - GM ta có:

xy+1=xy+14+14+14+14≥55√xy44.

Tương tự: yz+1≥55√yz44;zx+1≥55√zx44.

Do đó (xy+1)(yz+1)(zx+1)≥1255√(xyz)2412

⇒(xy+1)(yz+1)(zx+1)xyz≥1255√1412(xyz)3.

Mà xyz≤(x+y+z)327=18

Nên  (xy+1)(yz+1)(zx+1)xyz≥1255√83412=1255√1215=1258 

⇒P≥152.

\(\frac{27}{3\sqrt{3x-2}+6}+\frac{8+4x-x^2}{x\sqrt{6-x}+4}\ge\frac{3}{2}+\frac{2x-14}{3\sqrt{6-x}+2}>0\)

Nên phần còn lại vô nghiệm

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

Áp dụng BĐT Cauchy=Schwarz ta có:

\(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\Rightarrow x+y+z\le\sqrt{3}\)

Ta lại có:\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\forall x,y,z\)

\(\Leftrightarrow2\left(x^2+y^2+z^2\right)-2\left(xy+yz+zx\right)\ge0\)

\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+zx\)

\(\Rightarrow A\le\sqrt{3}+1\)

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

Em làm lại,cách này mà còn sai nữa thì em xin hàng ạ! Dù sao đi nữa cũng xin mọi người chịu khó góp ý giúp em để em càng ngày càng tiến bộ hơn nữa ạ! Thanks all !

*Tìm min

Đặt p = x + y + z; q = xy + yz + zx thì \(x^2+y^2+z^2=p^2-2q=1\Rightarrow q=\frac{p^2-1}{2}\)

Suy ra \(A=p+q=p+\frac{p^2-1}{2}=\frac{p^2+2p-1}{2}\)

\(=\frac{p^2+2p+1-2}{2}=\frac{\left(p+1\right)^2-2}{2}\ge-\frac{2}{2}=-1\)

Vậy giá trị nhỏ nhất của A là -1.

Dấu "=" xảy ra khi (x;y;z) = (0;0;-1) (chỗ này em không biết giải rõ thế nào nữa :v)

*Tìm max

Ta có BĐT sau: \(xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}\le x^2+y^2+z^2\)

Suy ra \(q\le\frac{p^2}{3}\le p^2-2q=1\) suy ra \(\hept{\begin{cases}q\le p^2-2q=1\\p^2\le3\left(p^2-2q\right)\end{cases}}\Leftrightarrow\hept{\begin{cases}q\le1\\p\le\sqrt{3\left(p^2-2q\right)}=\sqrt{3}\end{cases}}\)

Suy ra \(A=p+q\le\sqrt{3}+1\)

Ta có : 

\(x^2+y^2\ge2\sqrt{x^2y^2}=2xy\); \(y^2+z^2\ge2\sqrt{y^2z^2}=2yz\); \(x^2+z^2\ge2\sqrt{x^2z^2}=2xz\)

\(\Rightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\)\(\Rightarrow x^2+y^2+z^2\ge xy+yz+xz\)

hay \(xy+yz+xz\le1\)

Mặt khác : \(x^2+1\ge2x;y^2+1\ge2y;z^2+1\ge2z\)nên \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\)

hay \(2\left(x+y+z\right)\le4\)\(\Rightarrow x+y+z\le2\)

\(\Rightarrow A=x+y+z+xy+yz+xz\le2+1=3\)

hình như làm thế này sai thì phải