Nhờ mọi người giải ạ

Giải thích các bước giải:

Ta có

$M=\dfrac{1}a+\dfrac1{4b}+\dfrac1{16c}$

$\to M=\dfrac{16}{16a}+\dfrac{4}{16b}+\dfrac1{16c}$

$\to M=\dfrac{4^2}{16a}+\dfrac{2^2}{16b}+\dfrac{1^2}{16c}$

$\to M\ge\dfrac{(4+2+1)^2}{16a+16b+16c}$

$\to M\ge\dfrac{49}{16(a+b+c)}$

$\to M\ge\dfrac{49}{16}$

$\to GTNN_M=\dfrac{49}{16}$

Dấu = xảy ra khi $\dfrac{4}{16a}=\dfrac{2}{16b}=\dfrac1{16c}$

$\to \dfrac{1}{4a}=\dfrac{1}{8b}=\dfrac{1}{16c}$

$\to 4a=8b=16c$

$\to a=2b=4c$

$\to b=\dfrac12a, c=\dfrac14a$

Mà $a+b+c=1$

$\to a+\dfrac12a+\dfrac14a=1$

$\to a=\dfrac47\to b=\dfrac27, c=\dfrac17$