\(A=2xy+yz+xz\)

\(=2xy+y\left(4-2x-2y\right)+x\left(4-2x-2y\right)\)

\(=-2x^2-2xy+4x-2y^2+4y\)

\(=\left[-\left(x^2+2xy+y^2\right)+\dfrac{8}{3}\left(x+y\right)-\dfrac{16}{9}\right]-\left(x^2-\dfrac{4}{3}x+\dfrac{4}{9}\right)-\left(y-\dfrac{4}{3}y+\dfrac{4}{9}\right)+\dfrac{8}{3}\)\(=-\left(x+y-\dfrac{4}{3}\right)^2-\left(x-\dfrac{2}{3}\right)^2-\left(y-\dfrac{2}{3}\right)^2+\dfrac{8}{3}\le\dfrac{8}{3}\)

Vậy \(A_{max}=\dfrac{8}{3}\) tại \(\left\{{}\begin{matrix}x=y=\dfrac{2}{3}\\z=\dfrac{4}{3}\end{matrix}\right.\)

z = 4-2(x+y)

=> A= 2xy + y[4-2(x+y)] + x[4-2(x+y)]

=\(2xy+4y-2xy-2y^2+4x-2x^2-2xy\)

= \(-\left(y^2-4y+4\right)-\left(x^2-4x+4\right)-\left(y^2+2xy+x^2\right)+8\)

=\(8-\left[\left(y-2\right)^2+\left(x-2\right)^2-\left(y-x\right)^2\right]\le8\forall x,y\)

vậy GTLN của A là 8 khi x=y=2

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

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 

2x + 2y + z = 4(1)
A = 2xy + yz + xz(2)
(1) z=2c<=>x+y=2-c($)
(2)<=>2xy+2yc+2cx=A
A=2B<=>xy +(x+y).c=B
xy=B-c(2-c)
($:%)=> ton tai nghiem x,y
(c-2)^2≥4[B+c(c-2)]
c^2-4c+4≥4B+4c^2-8c
-3c^2+4c≥4B-4
-3(c^2-2.2/3c+4/9)≥4B-4-4/3
-3(c-2/3)^2≥4B-16/3
=> B≤4/3
A≤8/3
dang thuc khi c=2/3; z=1/3
x=y=2/3

A=2xy+yz+xzA=2xy+yz+xz

=2xy+y(4−2x−2y)+x(4−2x−2y)=2xy+y(4−2x−2y)+x(4−2x−2y)

=−2x2−2xy+4x−2y2+4y=−2x2−2xy+4x−2y2+4y

=[−(x2+2xy+y2)+83(x+y)−169]−(x2−43x+49)−(y−43y+49)+83=[−(x2+2xy+y2)+83(x+y)−169]−(x2−43x+49)−(y−43y+49)+83=−(x+y−43)2−(x−23)2−(y−23)2+83≤83=−(x+y−43)2−(x−23)2−(y−23)2+83≤83

Vậy Amax=83Amax=83 tại 

https://h.vn/hoi-dap/question/604792.html

Bn tham khảo tại đây nhé !

___G-Dragon___

mấy bạn chuyên toán giải giùm mk bài b) giùm ạ, mk đaq rất cần

\(P\le\frac{1}{2}\left(\Sigma\frac{1}{\sqrt{xy}}\right)\le\frac{\left(xy+yz+zx\right)^2}{6x^2y^2z^2}\le\frac{\left(x^2+y^2+z^2\right)^2}{6x^2y^2z^2}=\frac{3}{2}\)

dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=1\)

mình nhầm :) làm lại nhé

\(P\le\frac{1}{2}\left(\Sigma\frac{1}{\sqrt{xy}}\right)\le\frac{\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}{6xyz}\le\frac{xy+yz+zx}{2xyz}\le\frac{x^2+y^2+z^2}{2xyz}=\frac{3}{2}\)