\(P=\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\ge\sqrt{\dfrac{1}{2}\left(a+b\right)^2}+\sqrt{\dfrac{1}{2}\left(b+c\right)^2}+\sqrt{\dfrac{1}{2}\left(c+a\right)^2}\)
\(P\ge\dfrac{1}{\sqrt{2}}\left(a+b+b+c+c+a\right)=\sqrt{2}\left(a+b+c\right)=\sqrt{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)