MHB Find Min Value of a+b+c=1: $\sqrt {a^2+b^2}+\sqrt {b^2+c^2}+\sqrt {c^2+a^2}$

[Albert1]

$a>0,b>0,c>0 ,\,\, and \,\, a+b+c=1$

find $min(\sqrt {a^2+b^2}+\sqrt {b^2+c^2}+\sqrt {c^2+a^2} )$

 

[anemone]

$a>0,b>0,c>0 ,\,\, and \,\, a+b+c=1$

find $min(\sqrt {a^2+b^2}+\sqrt {b^2+c^2}+\sqrt {c^2+a^2} )$

My solution:

Spoiler

Note that $2(a^2+b^2)\ge (a+b)^2$ always holds for all real $a$ and $b$.

Therefore we have

$\begin{align*}\sqrt {a^2+b^2}+\sqrt {b^2+c^2}+\sqrt {c^2+a^2}&\ge \dfrac{2}{\sqrt{2}}\left(a+b+c\right)\\&\ge \dfrac{2(1)}{\sqrt{2}}\\&\ge \sqrt{2}\end{align*}$

Equality occurs when $a=b=c=\dfrac{1}{3}$.

 

[kaliprasad]

$a>0,b>0,c>0 ,\,\, and \,\, a+b+c=1$

find $min(\sqrt {a^2+b^2}+\sqrt {b^2+c^2}+\sqrt {c^2+a^2} )$

Spoiler

From cyclic symmetry we have $a = b = c = \frac{1}{3}$ is minumum of maximum
giving $min(\sqrt {a^2+b^2}+\sqrt {b^2+c^2}+\sqrt {c^2+a^2} )= \sqrt(2)$
it is minimum because at $(1,0,0)$ it is $2\sqrt(2)$

Note: while I was solving anemone beat me