Question

Cho x,y >0 tm 9x² + 4y² = 18

tim GTNN A= \(\frac{2}{x}+\frac{3}{y}+\frac{12}{3x+2y}\)

Answer 1

A = \(\frac{6}{3x}+\frac{6}{2y}+\frac{12}{3x+2y}=6.\left(\frac{1}{3x}+\frac{1}{2y}\right)+\frac{12}{3x+2y}\)

Áp dụng BĐT: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b};\)với a;b không âm

=> A \(\ge6.\frac{4}{3x+2y}+\frac{12}{3x+2y}=\frac{36}{3x+2y}\)

Mặt khác, (3x + 2y)² = (3x.1 + 2y.1)² \(\le\) (1² + 1²).(9x² + 4y²) = 2.18 = 36

=>  0< 3x + 2y \(\le\) 6 => \(\frac{36}{3x+2y}\ge\frac{36}{6}=6\)

=> A \(\ge\) 6.

Vậy Min A = 6 khi 3x = 2y => 18x² = 18 => x = 1 (do x > 0) => y = 3/2