Find Min Value of $(s-a)^3+(s-b)^3+(s-c)^3$ for $\triangle ABC$

In summary, the formula for finding the minimum value of $(s-a)^3+(s-b)^3+(s-c)^3$ for a given triangle is $3s(s-a)(s-b)(s-c)$, and to determine this value for a specific triangle, you can plug in the values of the side lengths into the formula. The minimum value represents the minimum possible sum of the cubes of the differences between the semi-perimeter $s$ and each side length, and it is always greater than or equal to 0 for any triangle. However, in the special case of an equilateral triangle, the minimum value will be equal to 0.
  • #1
Albert1
1,221
0
$a,b,c$ are lengths of $\triangle ABC$

if:

$(1) :s=\dfrac {a+b+c}{2}$, and

$(2) :$ the area of $\triangle ABC=1$

find the minimum value of $(s-a)^3+(s-b)^3+(s-c)^3$
 
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  • #2
My solution:

Applying the AM-GM inequality on the sum:
$(s-a)^3+(s-b)^3+(s-c)^3 \ge 3(s-a)(s-b)(s-c)\;\;\;(1)$
Using Herons formula (with area $A = 1$):
$\frac{1}{s}=(s-a)(s-b)(s-c)$
Inserting in $(1)$: $(s-a)^3+(s-b)^3+(s-c)^3 \ge \frac{3}{s}$The minimum of the sum is obtained, when $s = \frac{a+b+c}{2}$ is largest. Since the area of the triangle is fixed and $s$ is symmetric in $a,b, c$, it will occur when $a=b=c$, thus $s = \frac{3a}{2}$. Using Herons formula for this $s$- value yields:
$A = 1 = \frac{\sqrt{3}}{4}a^2 \Rightarrow a = \frac{2}{\sqrt[4]{3}}$,

and thus the minimum of the sum $(s-a)^3+(s-b)^3+(s-c)^3$ is:
\[\frac{3}{s}=\frac{2}{a}=2\cdot \frac{\sqrt[4]{3}}{2}=\sqrt[4]{3}.\]
 
Last edited:
  • #3
lfdahl said:
My solution:

Applying the AM-GM inequality on the sum:
$(s-a)^3+(s-b)^3+(s-c)^3 \ge 3(s-a)(s-b)(s-c)\;\;\;(1)$
Using Herons formula (with area $A = 1$):
$\frac{1}{s}=(s-a)(s-b)(s-c)$
Inserting in $(1)$: $(s-a)^3+(s-b)^3+(s-c)^3 \ge \frac{3}{s}$The minimum of the sum is obtained, when $s = \frac{a+b+c}{2}$ is largest. Since the area of the triangle is fixed and $s$ is symmetric in $a,b, c$, it will occur when $a=b=c$, thus $s = \frac{3a}{2}$. Using Herons formula for this $s$- value yields:
$A = 1 = \frac{\sqrt{3}}{4}a^2 \Rightarrow a = \frac{2}{\sqrt[4]{3}}$,

and thus the minimum of the sum $(s-a)^3+(s-b)^3+(s-c)^3$ is:
\[\frac{3}{s}=\frac{2}{a}=2\cdot \frac{\sqrt[4]{3}}{2}=\sqrt[4]{3}.\]
nice solution !
 

FAQ: Find Min Value of $(s-a)^3+(s-b)^3+(s-c)^3$ for $\triangle ABC$

What is the formula for finding the minimum value of $(s-a)^3+(s-b)^3+(s-c)^3$ for a given triangle?

The formula for finding the minimum value of $(s-a)^3+(s-b)^3+(s-c)^3$ is $3s(s-a)(s-b)(s-c)$, where $s$ is the semi-perimeter of the triangle.

How do you determine the minimum value of $(s-a)^3+(s-b)^3+(s-c)^3$ for a specific triangle?

To determine the minimum value of $(s-a)^3+(s-b)^3+(s-c)^3$ for a specific triangle, you can plug in the values of the triangle's side lengths into the formula $3s(s-a)(s-b)(s-c)$. The resulting value will be the minimum value for that triangle.

What does the minimum value of $(s-a)^3+(s-b)^3+(s-c)^3$ represent for a triangle?

The minimum value of $(s-a)^3+(s-b)^3+(s-c)^3$ represents the minimum possible sum of the cubes of the differences between the semi-perimeter $s$ of a triangle and each of its side lengths $a$, $b$, and $c$. It is a measure of the overall "compactness" of the triangle.

How can you prove that the minimum value of $(s-a)^3+(s-b)^3+(s-c)^3$ is always greater than or equal to 0 for any triangle?

Using the Triangle Inequality theorem, we know that for any triangle, $s-a$, $s-b$, and $s-c$ are all positive values. Since we are taking the sum of the cubes of these values, the minimum value must be greater than or equal to 0.

Are there any special cases where the minimum value of $(s-a)^3+(s-b)^3+(s-c)^3$ is equal to 0?

Yes, if the triangle is equilateral, with all three side lengths equal, the minimum value of $(s-a)^3+(s-b)^3+(s-c)^3$ will be equal to 0. This is because in an equilateral triangle, the semi-perimeter $s$ is equal to the side length $a$, so $s-a=0$. Similarly, $s-b=0$ and $s-c=0$, resulting in a minimum value of 0.

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