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

[Albert1]

$a,b,c$ 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 $(s-a)^3+(s-b)^3+(s-c)^3$

 

[lfdahl]

My solution:

Spoiler

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

 

[Albert1]

My solution:

Spoiler

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 !