\(\text{Ta có }:a^2+ab+b^2=\left(a^2+2ab+b^2\right)-ab\\ =\left(a+b\right)^2-ab\overset{BĐT\text{ }Cô-si}{\le}\left(a+b\right)^2-\frac{\left(a+b\right)^2}{4}=\frac{3}{4}\left(a+b\right)^2\\ \Rightarrow\sqrt{a^2+ab+b^2}\le\frac{\sqrt{3}}{2}\left(a+b\right)\)
Tương tự : \(\sqrt{b^2+bc+c^2}\le\frac{\sqrt{3}}{2}\left(b+c\right)\)
\(\sqrt{a^2+ac+c^2}\le\frac{\sqrt{3}}{2}\left(a+c\right)\\ \Rightarrow\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{a^2+ac+c^2}\\ \le\frac{\sqrt{3}}{2}\left(a+b\right)+\frac{\sqrt{3}}{2}\left(b+c\right)+\frac{\sqrt{3}}{2}\left(a+c\right)\\= \frac{\sqrt{3}}{2}\left(a+b+b+c+a+c\right)=\sqrt{3}\left(a+b+c\right)=3\sqrt{3}\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}a=b\\b=c\\a=c\\a+b+c=3\end{matrix}\right.\)
\(\Leftrightarrow a=b=c=1\)
Xin lỗi bạn nha. Tại mk không để ý.