Ta có:

4 A = ( x + y + z + t ) 2 ( x + y + z ) ( x + y ) x y z t ≥ 4 ( x + y + z ) t ( x + y + z ) ( x + y ) x y z t = 4 ( x + y + z ) 2 ( x + y ) x y z ≥ 4.4 ( x + y ) z ( x + y ) x y z = 16 ( x + y ) 2 x y ≥ 16.4 x y x y ≥ 64 ⇒ A ≥ 16

Đẳng thức xảy ra khi và chỉ khi  x + y + z + t = 2 x + y + z = t x + y = z x = y ⇔ x = y = 1 4 z = 1 2 t = 1

Áp dụng BĐT Cauchy, ta có:

4A = (x + y + z + t)²(x + y + z)(x + y)/xyzt

>= 4(x + y + z)t(x + y + z)(x + y)/xyzt

>= 4(x + y + z)²(x + y)/xyz >= 4 . 4(x + y)z(x + y)/xyz

>= 16(x + y)²/xy >= 16 . 4xy/xy >= 64

=> A >= 16

\(B\ge\dfrac{4\left(x+y+z\right)\left(x+y\right)}{\left(x+y\right)^2zt}=\dfrac{4\left(x+y+z\right)}{\left(x+y\right)zt}\ge\dfrac{16\left(x+y+z\right)}{\left(x+y+z\right)^2t}\)

\(B\ge\dfrac{16}{\left(x+y+z\right)t}\ge\dfrac{64}{\left(x+y+z+t\right)^4}=64\)

\(B_{min}=64\) khi \(\left(x;y;z;t\right)=\left(\dfrac{1}{8};\dfrac{1}{8};\dfrac{1}{4};\dfrac{1}{2}\right)\)

Áp dụng BĐT Cô si ta có :

+) \(x+y\ge2\sqrt{xy}\)

+) \(\left(x+y\right)+z\ge2\sqrt{\left(x+y\right)z}\)

+) \(\left(x+y+z\right)+t\ge2\sqrt{\left(x+y+z\right)t}\) 

Nhân từng vế với vế của các BĐT trên ta có :

\(\left(x+y\right)\left(x+y+z\right)\left(x+y+z+t\right)\ge8\sqrt{xyzt\left(x+y\right)\left(x+y+z\right)}\)

\(\Leftrightarrow2\left(x+y\right)\left(x+y+z\right)\ge8\sqrt{xyzt\left(x+y\right)\left(x+y+z\right)}\)

\(\Leftrightarrow\sqrt{\left(x+y\right)\left(x+y+z\right)}\ge4\sqrt{xyzt}\)

\(\Leftrightarrow\left(x+y\right)\left(x+y+z\right)\ge16xyzt\)

\(\Leftrightarrow B=\dfrac{\left(x+y\right)\left(x+y+z\right)}{xyzt}\ge16\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x+y=z\\x+y+z=t\\x+y+z+t=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=y=\dfrac{1}{4}\\z=\dfrac{1}{2}\\t=1\end{matrix}\right.\)

Vậy...

Ta có:

\(4A=\frac{\left(x+y+z+t\right)^2\left(x+y+z\right)\left(x+y\right)}{xyzt}\)

\(\ge\frac{4\left(x+y+z\right)t\left(x+y+z\right)\left(x+y\right)}{xyzt}\)

\(=\frac{4\left(x+y+z\right)^2\left(x+y\right)}{xyz}\ge\frac{16\left(x+y\right)z\left(x+y\right)}{xyz}\)

\(=\frac{16\left(x+y\right)^2}{xy}\ge\frac{64xy}{xy}=64\)

\(\Rightarrow A\ge16\)

Đấu = xảy ra khi \(t=2z=4x=4y=1\)

x;y;z;t >0 áp dụng bất đẳng thức Cô-si cho 2 số dương ta có :

=\(x+y\ge2\sqrt{xy}\)

=\(\left(x+y\right)+z\ge2\sqrt{\left(x+y\right)z}\)

=\(\left(x+y+z\right)+t\ge2\sqrt{\left(x+y+z\right)t}\)

nhân các vế tương ứng ta có:

\(\left(x+y\right)\left(x+y+z\right)\left(x+y+z+t\right)\ge8\sqrt{xyzt\left(x+y\right)\left(x+y+z\right)}\)

mà x+y+z+t=2

\(\left(x+y\right)\left(x+y+z\right)2\ge8\sqrt{xyzt\left(x+y\right)\left(x+y+z\right)}\)

=\(\sqrt{\left(x+y\right)\left(x+y+z\right)}\ge4\sqrt{xyzt}\)

=\(\left(x+y\right)\left(x+y+z\right)\ge16xyzt\)

\(\Rightarrow B=\frac{\left(x+y\right)\left(x+y+z\right)}{xyzt}\ge\frac{16xyzt}{xyzt}=16\)

vậy minB=16 khi\(\hept{\begin{cases}x=y\\x+y=z\\x+y+z=t\end{cases}};x+y+z+t=2\Rightarrow x=y=0.25;z=0.5;t=1\)

Ta có : \(2=\left[\left(x+y+z\right)+t\right]\ge4t\left(x+y+z\right)\)

\(\Rightarrow1\ge2t\left(x+y+z\right)\) (1)

Lại có : \(\left(x+y+z\right)^2=\left[\left(x+y\right)+z\right]^2\ge4z\left(x+y\right)\) (2)

\(\left(x+y\right)^2\ge4xy\) (3)

Nhân (1) , (2) , (3) theo vế được : 

\(\left(x+y\right)^2\left(x+y+z\right)^2\ge16xyzt\left(x+y\right)\left(x+y+z\right)\)

\(\Leftrightarrow\left(x+y\right)\left(x+y+z\right)\ge16xyzt\Leftrightarrow\frac{\left(x+y\right)\left(x+y+z\right)}{xyzt}\ge16\)

Suy ra Min B = 16 \(\Leftrightarrow\begin{cases}x+y+z=t\\x+y=z\\x=y\\x+y+z+t=2\end{cases}\)  \(\Leftrightarrow\begin{cases}x=y=\frac{1}{4}\\z=\frac{1}{2}\\t=1\end{cases}\)