Đáp án: $B\ge \dfrac{121}{108}$
Giải thích các bước giải:
Ta có:
$B=\dfrac1a+\dfrac1{4b}+\dfrac1{9c}$
$\to B=\dfrac{1^2}{a}+\dfrac{(\dfrac12)^2}b+\dfrac{(\dfrac13)^2}{c}$
$\to B\ge \dfrac{(1+\dfrac12+\dfrac13)^2}{a+b+c}$
$\to B\ge \dfrac{\dfrac{121}{36}}{3}$
$\to B\ge \dfrac{121}{108}$
Dấu = xay ra khi:
$\dfrac1a=\dfrac{\dfrac12}b=\dfrac{\dfrac13}c=\dfrac{1+\dfrac12+\dfrac13}{a+b+c}=\dfrac{11}{18}$
$\to a=\dfrac{18}{11}, b=\dfrac9{11}, c=\dfrac6{11}$
