MHB Find Min Value of a+b+c=1: $\sqrt {a^2+b^2}+\sqrt {b^2+c^2}+\sqrt {c^2+a^2}$

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Minimum Value
AI Thread Summary
The discussion focuses on 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 $a + b + c = 1$. Participants are engaged in deriving the minimum value, with one user noting that another, referred to as "anemone," arrived at a solution first. The problem emphasizes the relationships between the variables and the geometric interpretation of the expression. The conversation highlights the challenge of optimizing the expression while adhering to the given constraints. The thread showcases collaborative problem-solving in mathematical optimization.
Albert1
Messages
1,221
Reaction score
0
$a>0,b>0,c>0 ,\,\, and \,\, a+b+c=1$

find $min(\sqrt {a^2+b^2}+\sqrt {b^2+c^2}+\sqrt {c^2+a^2} )$
 
Mathematics news on Phys.org
Albert said:
$a>0,b>0,c>0 ,\,\, and \,\, a+b+c=1$

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

My solution:

Note that $2(a^2+b^2)\ge (a+b)^2$ always holds for all real $a$ and $b$.

Therefore we have

$\begin{align*}\sqrt {a^2+b^2}+\sqrt {b^2+c^2}+\sqrt {c^2+a^2}&\ge \dfrac{2}{\sqrt{2}}\left(a+b+c\right)\\&\ge \dfrac{2(1)}{\sqrt{2}}\\&\ge \sqrt{2}\end{align*}$

Equality occurs when $a=b=c=\dfrac{1}{3}$.
 
Albert said:
$a>0,b>0,c>0 ,\,\, and \,\, a+b+c=1$

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

From cyclic symmetry we have $a = b = c = \frac{1}{3}$ is minumum of maximum
giving $min(\sqrt {a^2+b^2}+\sqrt {b^2+c^2}+\sqrt {c^2+a^2} )= \sqrt(2)$
it is minimum because at $(1,0,0)$ it is $2\sqrt(2)$

Note: while I was solving anemone beat me
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top