Đáp án: $M=\dfrac{1}{3}$
Giải thích các bước giải:
Áp dụng bđt bunhia ta có:
$\sqrt{(a+2b)(a+2c)}\ge \sqrt{(\sqrt{a.a}+\sqrt{2b.2c})^2}=a+2\sqrt{bc}$
Tương tự ta chứng minh được
$\begin{cases}\sqrt{(b+2a)(b+2c)}\ge b+2\sqrt{ca}\\\sqrt{(c+2b)(c+2a)}\ge c+2\sqrt{ab}\end{cases}$
$\rightarrow \sqrt{(a+2b)(a+2c)}+\sqrt{(b+2a)(b+2c)}+\sqrt{(c+2b)(c+2a)}\ge a+2\sqrt{bc}+b+2\sqrt{ca}+c+2\sqrt{ab}\\$
$\rightarrow \sqrt{(a+2b)(a+2c)}+\sqrt{(b+2a)(b+2c)}+\sqrt{(c+2b)(c+2a)} \ge (\sqrt{a}+\sqrt{b}+\sqrt{c})^2\\$
$\rightarrow \sqrt{(a+2b)(a+2c)}+\sqrt{(b+2a)(b+2c)}+\sqrt{(c+2b)(c+2a)} \ge 3\\$
$\rightarrow \sqrt{(a+2b)(a+2c)}+\sqrt{(b+2a)(b+2c)}+\sqrt{(c+2b)(c+2a)} = 3\\$
$\leftrightarrow a=b=c=\dfrac{1}{3}$
$\rightarrow M=(2\sqrt{a}+3\sqrt{a}-4\sqrt{a})^2=a=\dfrac{1}{3}$
