Giải thích các bước giải:
Ta có:
$\sqrt{a(a+3b)}=\dfrac{1}{4}.2\sqrt{4a(a+3b)}\le \dfrac{1}{4}(4a+a+3b)=\dfrac{1}{4}(5a+3b)$
Tương tự
$\rightarrow \begin{cases}\sqrt{b(b+3c)}\le \dfrac{1}{4}(5b+3c)\\\sqrt{c(c+3a)}\le \dfrac{1}{4}(5c+3a) \end{cases}$
$\rightarrow \sqrt{a(a+3b)}+\sqrt{b(b+3c)}+\sqrt{c(c+3a)}\le \dfrac{1}{4}(5a+3b)+ \dfrac{1}{4}(5b+3c)+ \dfrac{1}{4}(5c+3a)$
$\rightarrow \sqrt{a(a+3b)}+\sqrt{b(b+3c)}+\sqrt{c(c+3a)}\le 2(a+b+c)$
$\rightarrow \dfrac{a+b+c}{\sqrt{a(a+3b)}+\sqrt{b(b+3c)}+\sqrt{c(c+3a)}}\ge \dfrac{1}{2}$
