Lời giải:
Đặt biểu thức đã cho là $A$
Ta có:
\(A=\sqrt{\frac{a}{b+c+2a}}+\sqrt{\frac{b}{a+c+2b}}+\sqrt{\frac{c}{a+b+2c}}\)
\(A=\sqrt{\frac{a}{(a+b)+(a+c)}}+\sqrt{\frac{b}{(b+c)+(b+a)}}+\sqrt{\frac{c}{(c+a)+(c+b)}}\)
Áp dụng BĐT AM-GM:
\(A\leq\sqrt{\frac{a}{2\sqrt{(a+b)(a+c)}}}+\sqrt{\frac{b}{2\sqrt{(b+c)(b+a)}}}+\sqrt{\frac{c}{2\sqrt{(c+a)(c+b)}}}\)
\(\Leftrightarrow A\leq \sqrt[4]{\frac{a^2}{4(a+b)(a+c)}}+\sqrt[4]{\frac{b^2}{4(b+c)(b+a)}}+\sqrt[4]{\frac{c^2}{4(c+a)(c+b)}}(*)\)
Tiếp tục áp dụng AM-GM:
\(\sqrt[4]{\frac{a^2}{4(a+b)(a+c)}}\leq \frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}+\frac{1}{2}+\frac{1}{2}\right)\)
\(\sqrt[4]{\frac{b^2}{4(b+c)(b+a)}}\leq \frac{1}{4}\left(\frac{b}{b+c}+\frac{b}{a+b}+\frac{1}{2}+\frac{1}{2}\right)\)
\(\sqrt[4]{\frac{c^2}{4(c+a)(c+b)}}\leq \frac{1}{4}\left(\frac{c}{c+a}+\frac{c}{c+b}+\frac{1}{2}+\frac{1}{2}\right)\)
Cộng theo vế kết hợp với $(*)$
\(\Rightarrow A\leq \frac{1}{4}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}+6.\frac{1}{2}\right)\)
\(\Leftrightarrow A\leq \frac{1}{4}.6=\frac{3}{2}\)
Vậy \(A_{\max}=\frac{3}{2}\Leftrightarrow a=b=c\)
