Biểu thức này chỉ có min, không có max

\(x^2+y^2+z^2=xy+xz+10yz\)

\(\Leftrightarrow\dfrac{3x^2}{4}+\left(\dfrac{x}{2}-y-z\right)^2=12yz\)

\(\Rightarrow12yz\ge\dfrac{3}{4}x^2\Rightarrow yz\ge\dfrac{x^2}{16}\)

\(\Rightarrow P\ge\dfrac{x^3}{2}-\dfrac{3x^3}{2yz}\ge\dfrac{x^3}{2}-\dfrac{3x^3}{\dfrac{x^2}{8}}=\dfrac{x^3}{2}-24x\)

Xét hàm \(f\left(x\right)=\dfrac{x^3}{2}-24x\) với \(x>0\Rightarrow f'\left(x\right)=\dfrac{3}{2}x^2-24=0\Rightarrow x=4\)

Từ BBT ta thấy \(\min\limits_{x>0}f\left(x\right)=f\left(4\right)=-64\)

\(\Rightarrow P_{min}=-64\) khi \(\left(x;y;z\right)=\left(4;1;1\right)\)

Vì x,y,z là các số nguyên dương

nên áp dụng bất đẳng thức Cauchy ta có :

\(x+y\ge2\sqrt{xy}\)(1)

\(y+z\ge2\sqrt{yz}\)(2)

\(z+x\ge2\sqrt{zx}\)(3)

Nhân (1), (2) và (3) theo vế ta có :

\(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge2\sqrt{xy}\cdot2\sqrt{yz}\cdot2\sqrt{zx}=8\sqrt{xy\cdot yz\cdot zx}=8\sqrt{x^2y^2z^2}=8\left|xyz\right|=8xyz\)

( do x,y,z là các số nguyên dương )

Đẳng thức xảy ra <=> x = y = z

=> đpcm

áp dụng BĐT AM-GM 

ta có \(x+y\ge2\sqrt{xy}\)

\(y+z\ge2\sqrt{yz}\)

\(z+x\ge2\sqrt{zx}\)

=>\(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}=8xyz\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}\Leftrightarrow x=y=z\left(ĐPCM\right)}\)

P/s : Mik ko chắc 

Áp dụng BĐT Cô - si , ta có : 

\(\frac{x+y}{2}\ge\sqrt{xy}\Rightarrow x+y\ge2\sqrt{xy}\left(1\right)\)

\(\frac{y+z}{2}\ge\sqrt{yz}\Rightarrow y+z\ge2\sqrt{yz}\left(2\right)\)

\(\frac{z+x}{2}\ge\sqrt{zx}\Rightarrow z+x\ge2\sqrt{zx}\left(3\right)\)

Từ \(\left(1\right);\left(2\right);\left(3\right)\)

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}\)

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge8.\sqrt{x^2y^2z^2}\)

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge8\left(\sqrt{xyz}\right)^2\)

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge8xyz\)

Dấu \("="\)

\(\Leftrightarrow x=y=z\)

\(\left(đpcm\right)\)

~ Ủng hộ nhé 

\(xy+yz+zx=8xyz\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=8\)

\(\Rightarrow\dfrac{8}{x}+\dfrac{8}{y}+\dfrac{8}{z}=64\)

Ta có: \(\dfrac{8}{x}+\dfrac{8}{y}+\dfrac{8}{z}\)

\(=\left(\dfrac{1}{x}+...+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\left(\dfrac{1}{y}+...+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{x}\right)+\left(\dfrac{1}{z}+...+\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{y}\right)\)

(sau dấu chấm là bốn số tương tự).

\(\ge^{Cauchy-Schwarz}\dfrac{8^2}{6x+y+z}+\dfrac{8^2}{6y+z+x}+\dfrac{8^2}{6z+x+y}\)

\(\Rightarrow64\ge\dfrac{8^2}{6x+y+z}+\dfrac{8^2}{6y+z+x}+\dfrac{8^2}{6z+x+y}\)

\(\Rightarrow\dfrac{1}{6x+y+z}+\dfrac{1}{6y+z+x}+\dfrac{1}{6z+x+y}\le1\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{3}{8}\)

Vậy \(Max\) của biểu thức đã cho là 1.

Bài toán không có lời giải vì không có số nguyên tố âm nên không có kết quả cho bài toán này

bn gõ bài trong công thức trực quan ik, khó nhìn lắm, ko làm đc

1) \(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)

\(=x^2y^3-x^3y^2+y^2z^3-y^3z^2-z^2x^2\left(z-x\right)\)

\(=\left(y^2z^3-x^3y^2\right)-\left(y^3z^2-x^2y^3\right)-z^2x^2\left(z-x\right)\)

\(=y^2\left(z^3-x^3\right)-y^3\left(z^2-x^2\right)-z^2x^2\left(z-x\right)\)

\(=y^2\left(z-x\right)\left(z^2+zx+x^2\right)-y^3\left(z-x\right)\left(z+x\right)-z^2x^2\left(z-x\right)\)

\(=\left(z-x\right)\left[y^2\left(z^2+zx+x^2\right)-y^3\left(z+x\right)-z^2x^2\right]\)

\(=\left(z-x\right)\left[\left(y^2z^2+xy^2z+x^2y^2\right)-\left(y^3z+xy^3\right)-z^2x^2\right]\)

\(=\left(z-x\right)\left(y^2z^2+xy^2z+x^2y^2-y^3z-xy^3-z^2x^2\right)\)

\(=\left(z-x\right)\left[\left(y^2z^2-y^3z\right)-\left(x^2z^2-x^2y^2\right)+\left(xy^2z-xy^3\right)\right]\)

\(=\left(z-x\right)\left[y^2z\left(z-y\right)-x^2\left(z^2-y^2\right)+xy^2\left(z-y\right)\right]\)

\(=\left(z-x\right)\left[y^2z\left(z-y\right)-x^2\left(z-y\right)\left(z+y\right)+xy^2\left(z-y\right)\right]\)

\(=\left(z-x\right)\left(z-y\right)\left[y^2z-x^2\left(z+y\right)+xy^2\right]\)

\(=\left(z-x\right)\left(z-y\right)\left(y^2z-x^2z-x^2y+xy^2\right)\)

\(=\left(z-x\right)\left(z-y\right)\left[\left(y^2z-x^2z\right)-\left(x^2y-xy^2\right)\right]\)

\(=\left(z-x\right)\left(z-y\right)\left[z\left(y^2-x^2\right)-xy\left(x-y\right)\right]\)

\(=\left(z-x\right)\left(z-y\right)\left[z\left(y-x\right)\left(y+x\right)+xy\left(y-x\right)\right]\)

\(=\left(z-x\right)\left(z-y\right)\left(y-x\right)\left[z\left(y+x\right)+xy\right]\)

\(=\left(z-x\right)\left(z-y\right)\left(y-x\right)\left(yz+xz+xy\right)\)

2) \(xyz-\left(xy+yz+xz\right)+\left(x+y+z\right)-1\)

\(=xyz-xy-yz-xz+x+y+z-1\)

\(=\left(xyz-xy\right)-\left(yz-y\right)-\left(xz-x\right)+\left(z-1\right)\)

\(=xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(z-1\right)\)

\(=\left(z-1\right)\left(xy-y-x+1\right)\)

\(=\left(z-1\right)\left[\left(xy-y\right)-\left(x-1\right)\right]\)

\(=\left(z-1\right)\left[y\left(x-1\right)-\left(x-1\right)\right]\)

\(=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)