A= 2xyz - xy - yz - zx +1 
= z(xy-1) - (xy-1) + zy(x-1) - z(x-1) 
= (z-1)(xy-1) + z(x-1)(y-1) 

Do x,y,z >1 nên A>0 suy ra đpcm

nguồn:Cho A= 2xyz - xy - yz - zx +1. Chứng minh A>0 với mọi x>1, y>1, z>1.

A= 2xyz - xy - yz - zx +1 
= z(xy-1) - (xy-1) + zy(x-1) - z(x-1) 
= (z-1)(xy-1) + z(x-1)(y-1) 

Do x,y,z >1 nên A>0 suy ra đpcm

Lời giải:
Áp dụng BĐT AM-GM:
$1=xy+yz+xz+2xyz\leq \frac{(x+y+z)^2}{3}+2.\frac{(x+y+z)^3}{27}$

$\Leftrightarrow 1\leq \frac{t^2}{3}+\frac{2t^3}{27}$ (đặt $x+y+z=t$)

$\Leftrightarrow 2t^3+9t^2-27\geq 0$

$\Leftrightarrow (t+3)^2(2t-3)\geq 0$

$\Leftrightarrow 2t-3\geq 0$
$\Leftrightarrow t\geq \frac{3}{2}$ hay $x+y+z\geq \frac{3}{2}$ (đpcm)

Dấu "=" xảy ra khi $x=y=z=\frac{1}{2}$

\(P=xy+yz+zx-2xyz=\left(xy+yz+zx\right)\left(x+y+z\right)-2xyz\)

\(P=xy\left(x+y\right)+yz\left(y+z\right)+zx\left(z+x\right)+xyz\ge0\)

Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(0;0;1\right)\) và hoán vị

Do vai trò của x;y;z là như nhau, ko mất tính tổng quát, giả sử \(z=min\left\{x;y;z\right\}\Rightarrow z\le\dfrac{1}{3}\)

\(P=xy\left(1-2z\right)+z\left(x+y\right)=xy\left(1-2z\right)+z\left(1-z\right)\)

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

\(P\le\dfrac{1+z^2-2z^3}{4}=\dfrac{1}{4}+\dfrac{z.z.\left(1-2z\right)}{4}\le\dfrac{1}{4}+\dfrac{1}{27.4}\left(z+z+1-2z\right)^3=\dfrac{7}{27}\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)

sai đề

Theo nguyên lý Dirichlet, trong 3 số x;y;z luôn có 2 số cùng phía so với \(\dfrac{1}{2}\)

Không mất tính tổng quát, giả sử đó là y và z 

\(\Rightarrow\left(y-\dfrac{1}{2}\right)\left(z-\dfrac{1}{2}\right)\ge0\Leftrightarrow yz-\dfrac{1}{2}\left(y+z\right)+\dfrac{1}{4}\ge0\)

\(\Leftrightarrow y+z-yz\le\dfrac{1}{2}+yz\)

Mặt khác từ giả thiết:

\(1-x^2=y^2+z^2+2xyz\ge2yz+2xyz\)

\(\Leftrightarrow\left(1-x\right)\left(1+x\right)\ge2yz\left(1+x\right)\)

\(\Leftrightarrow1-x\ge2yz\)

\(\Rightarrow yz\le\dfrac{1-x}{2}\)

Do đó:

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

\(\Rightarrow A\le-\dfrac{1}{2}x^2+\dfrac{1}{2}x+\dfrac{1}{2}=-\dfrac{1}{2}\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{8}\le\dfrac{5}{8}\)

\(A_{max}=\dfrac{5}{8}\) khi \(\left(x;y;z\right)=\left(\dfrac{1}{2};\dfrac{1}{2};\dfrac{1}{2}\right)\)

Ta có : \(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\)

\(=\left[xy\left(x+y\right)+xyz\right]+\left[yz\left(y+z\right)+xyz\right]+xz\left(x+z\right)\)

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

\(=y\left(x+y+z\right)\left(x+z\right)+xz\left(x+z\right)\)

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

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

\(x^{2019}+y^{2019}+z^{2019}=\left(x+y+z\right)^{2019}\)

Em xin lỗi, đây mới là đề đúng ạ !!

Hệ pt 3 ẩn mà chỉ có 2 pt à

Bài nào chả vậy

\(xy+yz+xz=2xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\Rightarrow\frac{1}{z^2}=\left(2-\frac{1}{x}-\frac{1}{y}\right)^2\)
Thay \(\frac{1}{z^2}=\left(2-\frac{1}{x}-\frac{1}{y}\right)^2\)vào pt \(\frac{2}{xy}=4+\frac{1}{z^2}\) ta đc pt:
\(\frac{2}{xy}=4+4+\frac{1}{x^2}+\frac{1}{y^2}-\frac{4}{x}-\frac{4}{y}+\frac{2}{xy}\)
\(\Leftrightarrow\left(\frac{1}{x^2}-\frac{4}{x}+4\right)+\left(\frac{1}{y^2}-\frac{4}{y}+4\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}-2\right)^2+\left(\frac{1}{y}-2\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x}=2\\\frac{1}{y}=2\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{2}\end{cases}}\Rightarrow z=-\frac{1}{2}}\)
Vậy nghiệm của hpt là (x;y;z)=(1/2;1/2;-1/2)