Đáp án: $A$

Giải thích các bước giải:

Kẻ $BD//MN, CE//MN, D\in AI, E\in AI$

$\to DB//CE$

$\to \dfrac{ID}{IE}=\dfrac{IB}{IC}=1$

$\to ID=IE$

$\to AD+AE=AI-ID+AI+IE=2AI$

Vì $MN//BD//CE$

$\to \dfrac{AB}{AM}+\dfrac{AC}{AN}=\dfrac{AD}{AO}+\dfrac{AE}{AO}=\dfrac{2AI}{AO}=\dfrac{2\cdot 2AO}{AO}=4$

Đặt $\dfrac{AB}{AM}=x\to x\in[1; 4)$

$\to \dfrac{AC}{AN}=4-x$

Ta có:
$\dfrac{S_{AMN}}{S_{ABC}}=\dfrac{AM}{AB}.\dfrac{AN}{AC}$

$\to \dfrac{S_{AMN}}{S_{ABC}}=\dfrac1x.\dfrac1{4-x}$

$\to \dfrac{S_{AMN}}{S_{ABC}}=\dfrac1{x(4-x)}\ge \dfrac1{\dfrac14(x+4-x)^2}=\dfrac14$

Mà $x\in[1; 4)$

$\to  \max\: \dfrac1{x(4-x)}=\dfrac13$ tại $x=1$

$\to \dfrac14\le\dfrac{S_{AMN}}{S_{ABC}}\le\dfrac13$

$\to \dfrac14\le\dfrac{S_{AMN}}{S}\le\dfrac13$

$\to \dfrac{S}4\le S_{AMN}\le\dfrac{S}3$