Lời giải:
Áp dụng BĐT Bunhiacopxky:
\((3a^2+b^2)(3+1)\geq (3a+b)^2\Rightarrow \sqrt{3a^2+b^2}\ge \frac{3a+b}{2}\)
\(\Rightarrow \frac{ab}{\sqrt{3a^2+b^2}+1}\leq \frac{2ab}{3a+b+2}\)
Thực hiện tương tự với các phân thức còn lại và cộng theo vế:
\(\Rightarrow Q\leq \frac{2ab}{3a+b+2}+\frac{2bc}{3b+c+2}+\frac{2ac}{3c+a+2}\)
\(\Leftrightarrow 3Q\leq \frac{6ab}{3a+b+2}+\frac{6bc}{3b+c+2}+\frac{6ac}{3c+a+2}\)
\(\Leftrightarrow 3Q\le 2b-\frac{2b^2+4b}{3a+b+2}+2c-\frac{2c^2+4c}{3b+c+2}+2a-\frac{2a^2+4a}{3c+a+2}\)
\(\Leftrightarrow 3Q\leq 6-\left(\frac{2b^2+4b}{3a+b+2}+\frac{2c^2+4c}{3b+c+2}+\frac{2a^2+4a}{3c+a+2}\right)(1)\)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{2b^2}{3a+b+2}+\frac{2c^2}{3b+c+2}+\frac{2a^2}{3c+a+2}\geq \frac{2(b+c+a)^2}{3a+b+2+3b+c+2+3c+a+2}=\frac{2(a+b+c)^2}{4(a+b+c)+6}=1(2)\)
Và:
\(\frac{4b}{3a+b+2}+\frac{4c}{3b+c+2}+\frac{4a}{3c+a+2}=4\left(\frac{b^2}{3ab+b^2+2b}+\frac{c^2}{3bc+c^2+2c}+\frac{a^2}{3ac+a^2+2a}\right)\)
\(\geq \frac{4(b+c+a)^2}{3ab+b^2+2b+3bc+c^2+3ac+a^2+2a}=\frac{4(a+b+c)^2}{(a+b+c)^2+2(a+b+c)+(ab+bc+ac)}\)
\(\geq \frac{4(a+b+c)^2}{(a+b+c)^2+2(a+b+c)+\frac{(a+b+c)^2}{3}}=2(3)\) (AM-GM)
Từ \((1); (2); (3)\Rightarrow 3Q\leq 6-(2+1)\Leftrightarrow 3Q\leq 3\Leftrightarrow Q\leq 1\)
Vậy Q(max) là $1$
Dấu bằng xảy ra khi \(a=b=c=1\)
