Đặt \(2z=a>0\)

Khi đó: \(\frac{1}{2}xya=x+y+a\)

\(\Rightarrow\frac{1}{2}=\frac{1}{xy}+\frac{1}{xa}+\frac{1}{ya}\)

Vì vai trò của 3 biến x,y,a là như nhau nên không mất tổng quát g/s: \(1\le x\le y\le a\)

Khi đó \(\frac{1}{2}=\frac{1}{xy}+\frac{1}{xa}+\frac{1}{ya}\le\frac{1}{x^2}+\frac{1}{x^2}+\frac{1}{x^2}=\frac{3}{x^2}\)

\(\Rightarrow x^2\le6\Rightarrow x\in\left\{1;2\right\}\)

Nếu x = 1 : \(yz=1+y+2z\)

\(\Leftrightarrow\left(yz-y\right)-\left(2z-2\right)=3\)

\(\Leftrightarrow\left(y-2\right)\left(z-1\right)=3\)

Xét PT ước nguyên dương khá dễ

Tương tự nếu x = 2 : 

\(2yz=2+y+2z\)

\(\Leftrightarrow\left(2yz-y\right)-\left(2z-1\right)=3\)

\(\Leftrightarrow\left(2z-1\right)\left(y-1\right)=3\)

Đến đây thì mình nghĩ chắc bạn cũng có thể tự giải được rồi!

bạn ơi đề khó nhìn vậy  

\(gt\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)

\(P=\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)

\(=\dfrac{1}{xyz}\left(x\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}+y\sqrt{\dfrac{5}{4}\left(x+z\right)^2+\dfrac{3}{4}\left(x-z\right)^2}+z\sqrt{\dfrac{5}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2}\right)\)

\(\ge\dfrac{1}{xyz}\left[x.\dfrac{\sqrt{5}\left(z+y\right)}{2}+y.\dfrac{\sqrt{5}\left(x+z\right)}{2}+z.\dfrac{\sqrt{5}\left(x+y\right)}{2}\right]\)

\(=\dfrac{\sqrt{5}\left(z+y\right)}{2yz}+\dfrac{\sqrt{5}\left(x+z\right)}{2xz}+\dfrac{\sqrt{5}\left(x+y\right)}{2xy}\)

\(=\dfrac{\sqrt{5}}{3}\left(1+1+1\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{\sqrt{5}}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2=\dfrac{\sqrt{5}}{3}\) (bunhia)

Dấu = xảy ra khi \(x=y=z=9\)

 Thấy : \(\sqrt{2y^2+yz+2z^2}=\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)>0\) 

CMTT : \(\sqrt{2x^2+xz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\)  ; \(\sqrt{2y^2+xy+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\) 

Suy ra : \(P\ge\dfrac{1}{xyz}.\dfrac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]\)

\(\Rightarrow P\ge\sqrt{5}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) 

Ta có : \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=\sqrt{xyz}\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\) 

Mặt khác :   \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2}{3}=\dfrac{1}{3}\)

Suy ra : \(P\ge\dfrac{\sqrt{5}}{3}\)

" = " \(\Leftrightarrow x=y=z=9\)

ap dung bdt \(x^{m+n}+y^{m+n}\ge x^my^n+x^ny^m\)  (bn tu cm )

\(\Rightarrow x^7+y^7=x^{3+4}+y^{3+4}\ge x^3y^4+x^4y^3\)

\(\Rightarrow\frac{x^2y^2}{x^2y^2+x^7+y^7}\le\frac{x^2y^2}{x^2y^2\left(1+xy^2+x^2y\right)}=\frac{1}{1+x^2y+y^2x}=\frac{1}{xyz+x^2y+y^2x}=\frac{1}{xy\left(x+y+z\right)}=\)

=\(\frac{z}{xyz\left(x+y+z\right)}=\frac{z}{x+y+z}\)

ttu \(P\le\frac{x+y+z}{x+y+z}=1\) đầu = xảy ra khi x=y=z=1

Không mất tính tổng quát giả sử : 0 < x\(\le\)y\(\le\)z.

Ta có: xyz = 2(x + y + z ) \(\le\)2 ( z + z + z ) = 6 z

Và xy = 2 ( x + y + z ) : z 

=> xyz \(\le\)6z

=> xy \(\le\)6

vì x, y là số nguyên dương

=> xy \(\in\){1; 2; 3; 4; 5; 6} với x\(\le\)y

+) TH1 : xy = 1 => x = y = 1

=> z = 2 ( 2 + z ) => z = 4 + 2z => z = -4 loại

+) TH2: xy = 2 => x = 1; y = 2 

=> 2 z = 2 ( 1 + 2 + z )  => 0z = 6 loại

+) TH3: xy = 3 => x = 1; y = 3

=> 3z = 2 ( 1 + 3 + z ) => z = 8  ( thỏa mãn )

+) Th4: xy = 4 => x =2 ; y = 2 hoặc x = 1; y =4

Với x =2; y = 2 => 4z =2 (  4+ z)  => z = 4 ( thỏa mãn )

Với x = 1; y = 4; => 4z = 2 ( 5 + z ) => z = 5 ( thỏa mãn)

Em làm tiếp nhé!

A=x^3 +y^3 +z^3+ 2(x/y+z  +y/z+x  +z/x+y)  \(\ge x^3+y^3+z^3+2.\frac{3}{2}\)  (bạn vào tìm BĐT nesbit là sẽ cm cái đằng sau >= 3/2)

Áp dụng cô si \(x^3+y^3+z^3\ge3xyz=3\)

===> A\(\ge3+3=6\) khi x=y=z=1

Ta có \(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\left(x,y,z>0\right)\).

\(\Leftrightarrow\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}=1\).

\(P=\frac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+y^2}\right)\)\(\left(x,y,z>0\right)\).

Ta có: 

\(\sqrt{2y^2+2yz+2z^2}=\sqrt{\frac{5}{4}\left(y^2+2yz+z^2\right)+\frac{3}{4}\left(y^2-2yz+z^2\right)}\)

\(=\sqrt{\frac{5}{4}\left(y+z\right)^2+\frac{3}{4}\left(y-z\right)^2}\).

Ta có:

\(\frac{3}{4}\left(y-z\right)^2\ge0\forall y;z>0\).

\(\Leftrightarrow\frac{3}{4}\left(y-z\right)^2+\frac{5}{4}\left(y+z\right)^2\ge\frac{5}{4}\left(y+z\right)^2\forall y;z>0\).

\(\Rightarrow\sqrt{\frac{3}{4}\left(y-z\right)^2+\frac{5}{4}\left(y+z\right)^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right)\forall y,z>0\).

\(\Leftrightarrow\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right)\forall y;z>0\).

\(\Leftrightarrow x\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}x\left(y+z\right)\forall x;y;z>0\left(1\right)\).

Chứng minh tương tự, ta được:

\(y\sqrt{2x^2+xz+2z^2}\ge\frac{\sqrt{5}}{2}y\left(x+z\right)\forall x;y;z>0\left(2\right)\).

Chứng minh tương tự, ta được:

\(z\sqrt{2x^2+xy+2y^2}\ge\frac{\sqrt{5}}{2}z\left(x+y\right)\forall x;y;z>0\left(3\right)\).

Từ \(\left(1\right),\left(2\right),\left(3\right)\), ta được:

\(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+2y^2}\)\(\ge\)\(\frac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]=\sqrt{5}\left(xy+yz+zx\right)\).

\(\Leftrightarrow\frac{1}{xyz}\left(x\sqrt{2y^2+yz+z^2}+y\sqrt{2z^2+zx+2x^2}+z\sqrt{2x^2+xy+2y^2}\right)\)\(\ge\)\(\frac{\sqrt{5}\left(xy+yz+zx\right)}{xyz}=\sqrt{5}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\).

\(\Leftrightarrow P\ge\frac{\sqrt{5}}{3}.3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{\sqrt{5}}{3}\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\)

\(\left(4\right)\).

Vì \(x,y,z>0\)nên áp dụng bất đẳng thức Bu-nhi-a-cốp-xki, ta được:
\(\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\)\(\left(1.\frac{1}{\sqrt{x}}+1.\frac{1}{\sqrt{y}}+1.\frac{1}{\sqrt{z}}\right)^2\).

\(\Leftrightarrow\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2=1^2=1\)

(vì\(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}=1\)).

\(\Leftrightarrow\frac{\sqrt{5}}{3}\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\frac{\sqrt{5}}{3}\)\(\left(5\right)\).

Từ \(\left(4\right)\)và \(\left(5\right)\), ta được:

\(P\ge\frac{\sqrt{5}}{3}\).

Dấu bằng xảy ra.

\(\Leftrightarrow\hept{\begin{cases}x=y=z>0\\\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\end{cases}}\Leftrightarrow x=y=z=9\).

Vậy \(minP=\frac{\sqrt{5}}{3}\Leftrightarrow x=y=z=9\).