Bài này chỉ tìm min chứ không tìm được max bạn nhé.

Áp dụng BĐT AM-GM:

\((3-2\sqrt{2})x^2+4y^2\geq 2\sqrt{4(3-2\sqrt{2})x^2y^2}=4(\sqrt{2}-1)xy\)

\((3-2\sqrt{2})z^2+4y^2\geq 4(\sqrt{2}-1)yz\)

\((-2+2\sqrt{2})x^2+(-2+2\sqrt{2})z^2\geq 2\sqrt{(-2+2\sqrt{2})^2x^2z^2}=4(\sqrt{2}-1)xz\)

Cộng theo vế và rút gọn thu được:

\(A\geq 4(\sqrt{2}-1)(xy+yz+xz)=4(\sqrt{2}-1)\)

Vậy $A_{\min}=4(\sqrt{2}-1)$

Bài này chỉ tìm min chứ không tìm được max bạn nhé.

Áp dụng BĐT AM-GM:

\((3-2\sqrt{2})x^2+4y^2\geq 2\sqrt{4(3-2\sqrt{2})x^2y^2}=4(\sqrt{2}-1)xy\)

\((3-2\sqrt{2})z^2+4y^2\geq 4(\sqrt{2}-1)yz\)

\((-2+2\sqrt{2})x^2+(-2+2\sqrt{2})z^2\geq 2\sqrt{(-2+2\sqrt{2})^2x^2z^2}=4(\sqrt{2}-1)xz\)

Cộng theo vế và rút gọn thu được:

\(A\geq 4(\sqrt{2}-1)(xy+yz+xz)=4(\sqrt{2}-1)\)

Vậy $A_{\min}=4(\sqrt{2}-1)$

\(\sqrt{2x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\le\dfrac{1}{2}\left(x+y+x+z\right)=\dfrac{1}{2}\left(2x+y+z\right)\)

Tương tự: \(\sqrt{2y+xz}\le\dfrac{1}{2}\left(x+2y+z\right)\) ; \(\sqrt{2z+xy}\le\dfrac{1}{2}\left(x+y+2z\right)\)

Cộng vế:

\(P\le\dfrac{1}{2}\left(4x+4y+4z\right)=4\)

\(P_{max}=4\) khi \(x=y=z=\dfrac{2}{3}\)

P = \(1.\sqrt{2x+yz}+1.\sqrt{2y+xz}+1.\sqrt{2z+xy}\)

\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(2x+yz+2y+xz+2z+xy\right)}\)

\(=\sqrt{3.\left(4+xy+yz+zx\right)}\)

Đã biết x² + y² + z² \(\ge\)xy + yz + zx

=> xy + yz + zx \(\le\dfrac{\left(x+y+z\right)^2}{3}\)

Khi đó \(P\le\sqrt{3\left(4+xy+yz+zx\right)}\le\sqrt{3\left[4+\dfrac{\left(x+y+z\right)^2}{3}\right]}\)

= 4 

Dấu "=" xảy ra <=> x = 2/3 

\(\sqrt{2x+yz}=\sqrt{\left(x+y+z\right)x+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\le\dfrac{x+2y+z}{2}\\ \Leftrightarrow P=\sum\sqrt{2x+yz}\le\dfrac{x+2y+z+2x+y+z+x+y+2z}{2}=\dfrac{4\left(x+y+z\right)}{2}=2\cdot2=4\)

Dấu \("="\Leftrightarrow x=y=z=\dfrac{2}{3}\)

P = \(1.\sqrt{2x+yz}+1.\sqrt{2y+xz}+1.\sqrt{2z+xy}\)

\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(2x+yz+2y+xz+2z+xy\right)}\)(Bunyacovski)

\(=\sqrt{3\left[4+\left(xy+yz+zx\right)\right]}\)

\(\le\sqrt{3.\left[4+\dfrac{\left(x+y+z\right)^2}{3}\right]}=\sqrt{3.\left(4+\dfrac{4}{3}\right)}\) = 4

Dấu "=" xảy ra <=> x = y = z = 2/3 

\(P=\sqrt{x\left(x+y+z\right)+yz}+\sqrt{y\left(x+y+z\right)+xz}+\sqrt{z\left(x+y+z\right)+xy}\)

\(P=\sqrt{\left(x+y\right)\left(x+z\right)}+\sqrt{\left(x+y\right)\left(y+z\right)}+\sqrt{\left(x+z\right)\left(y+z\right)}\)

\(P\le\dfrac{1}{2}\left(x+y+x+z\right)+\dfrac{1}{2}\left(x+y+y+z\right)+\dfrac{1}{2}\left(x+z+y+z\right)\)

\(P\le2\left(x+y+z\right)=2\)

\(P_{max}=2\) khi \(x=y=z=\dfrac{1}{3}\)

Ta có:
P=\(\left(X^2+y^2+z^2+2xyz\right)-\left(X^2+y^2+z^2+4xyz-xy-yz-xz\right)\) xz)
  = 1-\(\left(x^2+y^2+z^2+4xyz-xy-yz-xz\right)\)
=> P \(\le\)1
Vậy MaxP=1 

cô si cho gt