Đáp án:
Giải thích các bước giải:
$\quad \dfrac{a^2}{4}(BM^2 + CM^2)$
$=\dfrac{a^2}{4}[BM^2 + (BC - BM)^2]$
$=\dfrac{a^2}{4}\cdot \left[BM^2 + (a-BM)^2\right]$
$=\dfrac{a^2}{4}\cdot (BM^2 + a^2 - 2aBM + BM^2)$
$= \dfrac{a^2}{4}\cdot (2BM^2 - 2aBM + a^2)$
$= \dfrac{a^2}{4}\cdot 2\left(BM^2 - aBM + \dfrac{a^2}{2}\right)$
$=\dfrac{a^2}{2}\cdot \left(BM^2 - 2BM\cdot \dfrac a2 + \dfrac{a^2}{4} + \dfrac{a^2}{4}\right)$
$= \dfrac{a^2}{2}\cdot \left[\left(BM - \dfrac a2\right)^2 + \dfrac{a^2}{4}\right]$
$=\dfrac{a^2}{2}\cdot \left(BM - \dfrac a2\right)^2 + \dfrac{a^2}{2}\cdot \dfrac{a^2}{4}$
$= \dfrac{a^2}{2}\cdot \left(BM - \dfrac a2\right)^2 + \dfrac{a^4}{8}$
Ta có:
$\quad \left(BM - \dfrac a2\right)^2 \geqslant 0$
$\Leftrightarrow \dfrac{a^2}{2}\cdot \left(BM - \dfrac a2\right)^2 \geqslant 0$
$\Leftrightarrow \dfrac{a^2}{2}\cdot \left(BM - \dfrac a2\right)^2 + \dfrac{a^4}{8} \geqslant \dfrac{a^4}{8}$
Dấu $=$ xảy ra $\Leftrightarrow BM = \dfrac a2\Leftrightarrow M$ là trung điểm $BC$
