Cho a, b, c > 0 thỏa mãn $a+2b+3c\ge 20$  . Tìm GTNN: $a+b+c+\frac{3}{a}+\frac{9}{2b}+\frac{4}{c}$ .

Đặt A = $a+b+c+\frac{3}{a}+\frac{9}{2b}+\frac{4}{c}$

4A = $4a+4b+4c+\frac{12}{a}+\frac{36}{2b}+\frac{16}{c}$

= $a+2b+3c+3a+\frac{12}{a}+2b+\frac{36}{2b}+c+\frac{16}{c}$

Áp dụng bất đẳng thức AM – GM:

$\frac{12}{a}+3a\ge 2\sqrt{12.3}=12$

$2b+\frac{36}{2b}\ge 2\sqrt{36}=12$

$c+\frac{16}{c}\ge 2\sqrt{16}=8$

$\Rightarrow 4A\ge 20+12+12+8=52$

$\Rightarrow A\ge 13$

Dấu “=” xảy ra a = 2, b = 3, c = 4.