Đáp án + Giải thích các bước giải:
$A=a+b+c+\dfrac1a+\dfrac1b+\dfrac1c$
$A=4a+4b+4c-3a-3b-3c+\dfrac1a+\dfrac1b+\dfrac1c$
$A=\left(4a+\dfrac1a\right)+\left(4b+\dfrac1b\right)+\left(4c+\dfrac1c\right)-3a-3b-3c$
$A=\left(4a+\dfrac1a\right)+\left(4b+\dfrac1b\right)+\left(4c+\dfrac1c\right)-3(a+b+c)$
$A=\left(4a+\dfrac1a\right)+\left(4b+\dfrac1b\right)+\left(4c+\dfrac1c\right)-\dfrac{9}{2}$
Áp dụng BĐT Cô-si ta có:
$4a+\dfrac1a ≥ 2\sqrt{4a.\dfrac1a}=4$
$4b+\dfrac1b ≥ 2\sqrt{4b.\dfrac1b}=4$
$4c+\dfrac1c ≥ 2\sqrt{4c.\dfrac1c}=4$
$⇒ \left(4a+\dfrac1a\right)+\left(4b+\dfrac1b\right)+\left(4c+\dfrac1c\right)-\dfrac{9}{2}≥4+4+4-\dfrac92=12-\dfrac92=\dfrac{15}{2}$
Dấu $"="$ xảy ra khi $a=b=c=\dfrac12$
Vậy $\min A=\dfrac{15}{2}$ khi $a=b=c=\dfrac12$
