Cho 3 số thực dương a,b,c thỏa mãn a+2b+3c ≥ 20. Tìm GTNN của A= a+b+c +3/a+9/2b+4/c 22/09/2021 Bởi Maya Cho 3 số thực dương a,b,c thỏa mãn a+2b+3c ≥ 20. Tìm GTNN của A= a+b+c +3/a+9/2b+4/c
$S=a+b+c+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}$ $=>S=\bigg(\dfrac{3a}{4}+\dfrac{3}{a}\bigg)+\bigg(\dfrac{b}{2}+\dfrac{9}{2b}\bigg)+\bigg(\dfrac{c}{4}+\dfrac{4}{c}\bigg)+\dfrac{1}{4}(a+2b+3c)$ $=>S≥2\sqrt{\dfrac{3a}{4}.\dfrac{3}{a}}+2\sqrt{\dfrac{b}{2}.\dfrac{9}{2b}}+2\sqrt{\dfrac{c}{4}.\dfrac{4}{c}}+\dfrac{1}{4}.20$ $=>S≥13$ Dấu $”=”$ xảy ra $<=>\left\{ {\matrix{{a=2} \cr{b=3} \cr{c=4}} } \right.$ Vậy $S_{min}=13$ tại $(a;b;c)=(2;3;4)$ Bình luận
$S=a+b+c+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}$
$=>S=\bigg(\dfrac{3a}{4}+\dfrac{3}{a}\bigg)+\bigg(\dfrac{b}{2}+\dfrac{9}{2b}\bigg)+\bigg(\dfrac{c}{4}+\dfrac{4}{c}\bigg)+\dfrac{1}{4}(a+2b+3c)$
$=>S≥2\sqrt{\dfrac{3a}{4}.\dfrac{3}{a}}+2\sqrt{\dfrac{b}{2}.\dfrac{9}{2b}}+2\sqrt{\dfrac{c}{4}.\dfrac{4}{c}}+\dfrac{1}{4}.20$
$=>S≥13$
Dấu $”=”$ xảy ra $<=>\left\{ {\matrix{{a=2} \cr{b=3} \cr{c=4}} } \right.$
Vậy $S_{min}=13$ tại $(a;b;c)=(2;3;4)$