Question

Cho x,y,z >0 thỏa mãn x+y+z=3 Tìm min A = \(\frac{x+y}{xyz}\)

Answer 1

Ta có \(3=x+y+z=x+y+\frac{z}{2}+\frac{z}{2}\ge4\sqrt[4]{x.y.\frac{z^2}{4}}\)

=> \(xyz^2\le\frac{81}{64}\)

\(A=\frac{x+y}{xyz}\ge\frac{2\sqrt{xy}}{xyz}=\frac{2}{\sqrt{xyz^2}}\ge\frac{2}{\sqrt{\frac{81}{64}}}=\frac{16}{9}\)

MinA=16/9  khi \(x=y=\frac{3}{4};z=\frac{3}{2}\)