Question

cho x,y,z dương thỏa \(\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{2}{z}=0\)

tìm MIN T=\(\dfrac{x+z}{2x-z}+\dfrac{z+y}{2y-z}\)

Answer 1

Theo đề thì:\(\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{2}{z}=0\)

\(\Leftrightarrow xz+yz-2xy=0\)

Cũng từ \(\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{2}{z}=0\)

\(\Leftrightarrow\dfrac{2}{z}=\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\)

\(\Leftrightarrow z\le\sqrt{xy}\)

\(\Leftrightarrow z^2\le xy\)

Quay lại bài toán ta có:

\(T=\dfrac{x+z}{2x-z}+\dfrac{z+y}{2y-z}=\dfrac{2z^2-6xy-\left(xz+yz-2xy\right)}{-z^2+2\left(xz+yz-2xy\right)}\)

\(=\dfrac{6xy-2z^2}{z^2}\ge\dfrac{6xy-2xy}{xy}=4\)

Vậy GTNN là T = 4 khi x = y = z = 1