Find Min Value: $a,b,c>0$ with $a+b+c=k$

[Albert1]

$a,b,c>0$

$a+b+c=k$

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

 

[juantheron]

My Solution:

Spoiler

Given $$a+b+c = k$$ and $$a,b,c>0$$

Now we can write $$\sqrt{a^2+b^2} = \left|a+ib\right|$$ and $$\sqrt{b^2+c^2} = \left|b+ic\right|$$ and $$\sqrt{c^2+a^2} = \left|c+ia\right|$$

Where $$i=\sqrt{-1}$$ So Using Triangle Inequality of Complex number

$$\left|a+ib\right|+\left|b+ic\right|+\left|c+ia\right|\geq \left|\left(a+b+c\right)+i\left(b+c+a\right)\right| = \left|k+ik\right|=\sqrt{2}k$$

and equality hold when $$\displaystyle \frac{a}{b} = \frac{b}{c} = \frac{c}{a}$$

 

[Albert1]

My Solution:

Spoiler

Given $$a+b+c = k$$ and $$a,b,c>0$$

Now we can write $$\sqrt{a^2+b^2} = \left|a+ib\right|$$ and $$\sqrt{b^2+c^2} = \left|b+ic\right|$$ and $$\sqrt{c^2+a^2} = \left|c+ia\right|$$

Where $$i=\sqrt{-1}$$ So Using Triangle Inequality of Complex number

$$\left|a+ib\right|+\left|b+ic\right|+\left|c+ia\right|\geq \left|\left(a+b+c\right)+i\left(b+c+a\right)\right| = \left|k+ik\right|=\sqrt{2}k$$

and equality hold when $$\displaystyle \frac{a}{b} = \frac{b}{c} = \frac{c}{a}$$

nice solution !

 

[Albert1]

$a,b,c>0$

$a+b+c=k$

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

Spoiler

View attachment 3308