\(P=\sqrt{\left(a+b\right)^2+b^2}+\sqrt{\left(b+c\right)^2+c^2}+\sqrt{\left(a+c\right)^2+a^2}\)
\(\Leftrightarrow\sqrt{5}P=\sqrt{5\left[\left(a+b\right)^2+b^2\right]}+\sqrt{5\left[\left(b+c\right)^2+c^2\right]}+\sqrt{5\left[\left(a+c\right)^2+a^2\right]}\)
Có : \(\sqrt{5\left[\left(a+b\right)^2+b^2\right]}=\sqrt{\left[\left(a+b\right)^2+b^2\right]\left(2^2+1^2\right)}\ge2\left(a+b\right)+b\)
(BĐT Bunyakovski)
Tương tự được \(\sqrt{5}P\ge2\left(a+b\right)+b+2\left(b+c\right)+c+2\left(a+c\right)+a\)
\(\Leftrightarrow\sqrt{5}P\ge5\left(a+b+c\right)\Leftrightarrow P\ge\sqrt{5}\)
"=" khi \(\left\{{}\begin{matrix}\dfrac{a+b}{2}=b\\\dfrac{b+c}{2}=c\\\dfrac{a+c}{2}=a\\a+b+c=1\end{matrix}\right.\Leftrightarrow a=b=c=\dfrac{1}{3}\)
