Albert1
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$a,b,c>0$
$a+b+c=k$
find:$min(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})$
$a+b+c=k$
find:$min(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})$
The discussion centers around finding the minimum value of the expression $\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}$ under the constraints that $a, b, c > 0$ and $a + b + c = k$. The scope includes mathematical reasoning and optimization techniques.
The discussion does not show clear consensus, as multiple participants have proposed solutions without resolving which, if any, is definitive.
Details of the proposed solutions are not provided, leaving the discussion open to interpretation and further exploration of methods.
nice solution !jacks said:My Solution:
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 $$\displaystyle \frac{a}{b} = \frac{b}{c} = \frac{c}{a}$$
Albert said:$a,b,c>0$
$a+b+c=k$
find:$min(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})$