Optimizing Triangular Inequalities: Finding the Minimum of a Complex Expression

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SUMMARY

The minimum value of the expression $\sqrt{a^2-12a+40}+\sqrt{b^2-8b+20}+\sqrt{a^2+b^2}$ is established as 10 using the Triangle Inequality. The equality condition is satisfied when the ratios $\frac{6-a}{2}=\frac{2}{4-b}=\frac{a}{b}$ hold true. This analysis provides a clear pathway to optimize the given complex expression through geometric interpretations.

PREREQUISITES
  • Understanding of Triangle Inequality in geometry
  • Familiarity with algebraic manipulation of square roots
  • Knowledge of optimization techniques in mathematics
  • Basic skills in solving equations and inequalities
NEXT STEPS
  • Study the application of Triangle Inequality in various mathematical contexts
  • Explore advanced optimization techniques in calculus
  • Learn about geometric interpretations of algebraic expressions
  • Investigate the properties of inequalities and their proofs
USEFUL FOR

Mathematicians, students studying optimization problems, and anyone interested in advanced algebraic techniques and geometric interpretations.

anemone
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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 $\displaystyle \frac{6-a}{2}=\frac{2}{4-b} = \frac{a}{b}$

- - - Updated - - -
 
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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 $\displaystyle \frac{6-a}{2}=\frac{2}{4-b} = \frac{a}{b}$

- - - Updated - - -

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

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