The minimum value of the expression $(s-a)^3+(s-b)^3+(s-c)^3$ for triangle $\triangle ABC$ with side lengths $a$, $b$, and $c$, where $s=\frac{a+b+c}{2}$ and the area of the triangle equals 1, has been discussed. The solution provided indicates that the problem can be approached using properties of triangle inequalities and optimization techniques. The conclusion emphasizes that the minimum occurs under specific conditions related to the triangle's dimensions.
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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:
\[\frac{3}{s}=\frac{2}{a}=2\cdot \frac{\sqrt[4]{3}}{2}=\sqrt[4]{3}.\]