Question

Cho 4 số thực dương \(x;y;z;t\) thỏa mãn \(x+y+z+t=2\)

Tìm giá trị nhỏ nhất của \(A=\frac{\left(x+y+z\right)\left(x+y\right)}{xyzt}\)

Answer 1

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

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

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

\(A_{min}=16\) khi \(\left\{{}\begin{matrix}x+y+z+t=2\\x+y+z=t\\x+y=z\\x=y\end{matrix}\right.\) \(\Rightarrow\left(x;y;z;t\right)=...\)