Giải thích các bước giải:
$\begin{split}\dfrac{4a^2}{a-1}&=\dfrac{4a^2-4+4}{a-1}\\&=\dfrac{4(a-1)(a+1)+4}{a-1}=4(a+1)+\dfrac{4}{a-1}\\&=4(a-1)+\dfrac{4}{a-1}+8\\&\ge 2\sqrt{4(a-1).\dfrac{4}{a-1}}+8=16\end{split}$
$\begin{split}\dfrac{5b^2}{b-1}&=\dfrac{5b^2-5+5}{b-1}\\&=5(b+1)+\dfrac{5}{b-1}\\&=5(b-1)+\dfrac{5}{b-1}+10\\&\ge 2\sqrt{5(b-1).\dfrac{5}{b-1}}+10=20\end{split}$
$\begin{split}\dfrac{3c^2}{c-1}&=\dfrac{3c^2-3+3}{c-1}\\&=3(c+1)+\dfrac{3}{c-1}\\&=3(c-1)+\dfrac{3}{c-1}+6\\&\ge 2\sqrt{3(c-1).\dfrac{3}{c-1}}+6=12\end{split}$
$\rightarrow \dfrac{4a^2}{a-1}+\dfrac{5b^2}{b-1}+\dfrac{3c^2}{c-1}\ge 16+20+12=48$
$\rightarrow đpcm$
