nice solution !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:
\[\frac{3}{s}=\frac{2}{a}=2\cdot \frac{\sqrt[4]{3}}{2}=\sqrt[4]{3}.\]
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.
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.
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.
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.
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.