MHB Optimizing Triangular Inequalities: Finding the Minimum of a Complex Expression

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The discussion focuses on minimizing the expression $\sqrt{a^2-12a+40}+\sqrt{b^2-8b+20}+\sqrt{a^2+b^2}$ using the Triangle Inequality. It establishes that the minimum value is 10, achieved under specific conditions relating the variables a and b. The equality condition is given by the ratios $\frac{6-a}{2}=\frac{2}{4-b}=\frac{a}{b}$. Participants express appreciation for contributions and insights shared in the thread. The mathematical approach effectively demonstrates the application of inequalities in optimization problems.
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Find the minimum of $\sqrt{a^2-12a+40}+\sqrt{b^2-8b+20}+\sqrt{a^2+b^2}$.
 
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anemone said:
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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jacks said:
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!
 
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