Find Min Value: $a,b,c>0$ with $a+b+c=k$

[Albert1]

$a,b,c>0$

$a+b+c=k$

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

 

[juantheron]

My Solution:

Spoiler

Given \(\displaystyle a+b+c = k\) and \(\displaystyle a,b,c>0\)

Now we can write \(\displaystyle \sqrt{a^2+b^2} = \left|a+ib\right|\) and \(\displaystyle \sqrt{b^2+c^2} = \left|b+ic\right|\) and \(\displaystyle \sqrt{c^2+a^2} = \left|c+ia\right|\)

Where \(\displaystyle i=\sqrt{-1}\) So Using Triangle Inequality of Complex number

\(\displaystyle \left|a+ib\right|+\left|b+ic\right|+\left|c+ia\right|\geq \left|\left(a+b+c\right)+i\left(b+c+a\right)\right| = \left|k+ik\right|=\sqrt{2}k\)

and equality hold when \(\displaystyle \displaystyle \frac{a}{b} = \frac{b}{c} = \frac{c}{a}\)

 

[Albert1]

My Solution:

Spoiler

Given \(\displaystyle a+b+c = k\) and \(\displaystyle a,b,c>0\)

Now we can write \(\displaystyle \sqrt{a^2+b^2} = \left|a+ib\right|\) and \(\displaystyle \sqrt{b^2+c^2} = \left|b+ic\right|\) and \(\displaystyle \sqrt{c^2+a^2} = \left|c+ia\right|\)

Where \(\displaystyle i=\sqrt{-1}\) So Using Triangle Inequality of Complex number

\(\displaystyle \left|a+ib\right|+\left|b+ic\right|+\left|c+ia\right|\geq \left|\left(a+b+c\right)+i\left(b+c+a\right)\right| = \left|k+ik\right|=\sqrt{2}k\)

and equality hold when \(\displaystyle \displaystyle \frac{a}{b} = \frac{b}{c} = \frac{c}{a}\)

nice solution !

 

[Albert1]

$a,b,c>0$

$a+b+c=k$

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

Spoiler

View attachment 3308

 

[CellCoree]

I would first clarify that the given equation is a constraint on the values of $a, b,$ and $c$, where $k$ is a constant. This means that the values of $a, b,$ and $c$ must satisfy the constraint $a+b+c=k$.

Next, the expression to be minimized is the sum of three square roots, which can be rewritten as the sum of three terms: $\sqrt{a^2+b^2}, \sqrt{b^2+c^2},$ and $\sqrt{c^2+a^2}$. Using the properties of square roots, we can see that each term can be written as the hypotenuse of a right triangle with sides $a, b,$ and $c$, respectively.

Since we are looking for the minimum value of the expression, we can use the concept of the Pythagorean theorem to find the minimum value of each term. According to the Pythagorean theorem, the hypotenuse of a right triangle is always greater than or equal to the sum of its two legs. Therefore, the minimum value of each term is when one of the legs is equal to zero, resulting in a minimum value of $\sqrt{a^2+b^2}=a$.

Therefore, the minimum value of the entire expression is achieved when $a=0$, which also satisfies the constraint $a+b+c=k$. This means that the minimum value of the expression is simply $a$, which is equal to $0$. In other words, the minimum value of $\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}$ is $0$, and it is achieved when $a=b=c=0$.

In conclusion, the minimum value of the given expression is $0$, and it is achieved when $a=b=c=0$, satisfying the constraint $a+b+c=k$.