Finding the minimum of √(a²-12a+40)+√(b²-8b+20)+√(a²+b²)

[anemone]

Find the minimum of $\sqrt{a^2-12a+40}+\sqrt{b^2-8b+20}+\sqrt{a^2+b^2}$.

 

[juantheron]

Find the minimum of $\sqrt{a^2-12a+40}+\sqrt{b^2-8b+20}+\sqrt{a^2+b^2}$.

Using Triangle Inequality::

$\displaystyle \sqrt{(6-a)^2+2^2}+\sqrt{2^2+(4-b)^2}+\sqrt{a^2+b^2}\geq \sqrt{(6-a+2+a)^2+(2+4-b+b)^2} = 10$

and equality hold when $\displaystyle \frac{6-a}{2}=\frac{2}{4-b} = \frac{a}{b}$

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[anemone]

Using Triangle Inequality::

$\displaystyle \sqrt{(6-a)^2+2^2}+\sqrt{2^2+(4-b)^2}+\sqrt{a^2+b^2}\geq \sqrt{(6-a+2+a)^2+(2+4-b+b)^2} = 10$

and equality hold when $\displaystyle \frac{6-a}{2}=\frac{2}{4-b} = \frac{a}{b}$

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Bravo, jacks!(Yes) And thanks for participating!