Áp dụng BĐT: \(x^2+y^2\ge\frac{1}{2}\left(x+y\right)^2\)
\(\Rightarrow VT\ge\sqrt{\frac{1}{2}\left(a+b\right)^2}+\sqrt{\frac{1}{2}\left(b+c\right)^2}+\sqrt{\frac{1}{2}\left(c+a\right)^2}\)
\(VT\ge\frac{1}{\sqrt{2}}\left(2a+2b+2c\right)=\sqrt{2}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)