Đáp án:
P+5= $\frac{3(b+c)}{2a}$ +$\frac{4a+3c}{3b}$ +$\frac{12(b-c)}{2a+3c}$ +4+1
P+5= $\frac{3b}{2a}$ +$\frac{3c}{2a}$+$\frac{2a}{3a}$+$\frac{2a}{3b}$+$\frac{3c}{3b}$+$\frac{4(2a+3b)}{2a+3c}$
P+5= $\frac{3b}{2a}$ +$\frac{2a}{3b}$+$\frac{3c}{2a}$+$\frac{3c}{3b}$+$\frac{2a}{3b}$+$\frac{2a}{2a}$+$\frac{4(2a+3b)}{2a+3c}$
áp dụng BĐT cosi $\frac{3b}{2a}$ +$\frac{2a}{3b}$≥2
$\frac{3c}{2a}$+$\frac{3c}{3b}$=3c($\frac{1}{2a}$+$\frac{1}{3b}$)
áp dụng BĐT svacso
$\frac{1}{2a}$+$\frac{1}{3b}$≥$\frac{4}{2a+3b}$
=> $\frac{3c}{2a}$+$\frac{3c}{3b}$=>$\frac{4.3c}{2a+3b}$
$\frac{2a}{3b}$+$\frac{2a}{2a}$=2a($\frac{1}{3b}$+$\frac{1}{2a}$)
$\frac{1}{3b}$+$\frac{1}{2a}$≥$\frac{4}{3b}$
=> $\frac{2a}{2a}$+$\frac{2a}{3b}$≥$\frac{4.2a}{3b+2a}$
=> P+5= $\frac{3b}{2a}$ +$\frac{2a}{3b}$+$\frac{3c}{2a}$+$\frac{3c}{3b}$+$\frac{2a}{3b}$+$\frac{2a}{2a}$+$\frac{4(2a+3b)}{2a+3c}$ ≥2+$\frac{4.2a}{3b}$+$\frac{4.3c}{2a+3b}$+$\frac{4(2a+3b)}{2a+3c}$
=2+4($\frac{2a}{3b+2a}$+$\frac{3c}{2a+3b}$+$\frac{2a+3b}{2a+3c}$)
=2+4($\frac{2a+3c}{3b+2a}$+$\frac{2a+3b}{2a+3c}$)
áp dụng BĐT co si $\frac{2a+3c}{3b+2a}$+$\frac{2a+3b}{2a+3c}$≥2
=> VT≥2+8=10
=> P+5≥10
=> P≥5
dấu"=' xảy ra <=> 2a=3b=3c