Áp dụng holder ta có:

\(\left(1+1+1\right)\left(x^2y+y^2z+z^2x\right)\left(xy^2+yz^2+zx^2\right)\)

\(\ge\left(\sqrt[3]{x^4yz}+\sqrt{y^4zx}+\sqrt{z^4xy}\right)^3=xyz\left(x+y+z\right)^3\)

Dạo này bận lắm nên cũng lười luôn nên thông cảm.

Bài này làm được theo 1 cách khác nhưng phải áp dụng 2 lần bđt

lần 1 dùng bđt Schur

lần 2 dùng AM-GM

Ta có:

\(xy+yz+zx=-5;xz=-5\)

\(\Rightarrow xy+yz=0\)

\(\Rightarrow y\left(x+z\right)=0\)

\(\Rightarrow\orbr{\begin{cases}y=0\\x+z=0\end{cases}}\)

Nếu \(y=0\) ta có:

\(x+0+z=2\Rightarrow x+z=2\)

\(A=x^3+y^3+z^3=\left(x+z\right)\left[\left(x+z\right)^2-3xz\right]+y^3=2\cdot\left(2^2+3\cdot5\right)+0=38\)

Nếu \(x+z=0\Rightarrow y=2\),ta có:

\(A=x^3+y^3+z^3=\left(x+z\right)\left[\left(x+z\right)^2-3xz\right]+y^3=8\)

Vậy \(A=8\left(h\right)A=38\)

cho minh sua lai la xyz=-5 nha

Ta có :

\(x^3+x^2z+y^2z-xyz+y^3\)

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

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

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

\(=0\left(x^2-xy+y^2\right)\)

\(=0\left(ĐPCM\right)\)

Lời giải:

Ta có:

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

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

\(=(x^2-xy+y^2)(x+y+z)=(x^2-xy+y^2).0=0\)

Ta có đpcm.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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