TITLE: The minimum value of $\frac{a^3 + b^3 + c^3 }{\sqrt{a^4 + b^4 + c^4 }}$ . When $a^2 + b^2 + c^2 = 1 $
QUESTION [5 upvotes]: Asume $a, b, c $ is non-negative real.
I got above equation at this situation ;
$\vec {x}= (a, b, c)$ , 
$\vec {y} = (a^2 , b^2 , c^2 ) $
$$ cos \phi = \frac{ \vec x \cdot \vec {y}} { \Vert {\vec {x} \Vert} \times\Vert {\vec {y} \Vert}} = \frac{ a^3 + b^3 + c^3 } { \sqrt{ a^4 + b^4 + c^4 }}$$
In this case I want to know the Maximum value of $\phi$ .
But I can't compute. 
I want your help.

REPLY [3 votes]: I think that Macavity gave a good answer. 

I would add a remark which could hide a geometrical property.

EDIT. The minimum of the function $f:a\rightarrow \dfrac{a^3+2}{\sqrt{(a^2+2)(a^4+2)}}$ is reached in $a_0$ satisfying $a_0^4-2a_0^3-4a_0+2=0$. It is easy to see that $a_0$ can be constructed with rule and compass over $\mathbb{Q}(3^{1/3})$ ($a_0$ is constructible if we have the length $3^{1/3}$). We have to normalize $a_0,b_0=c_0=1$. An admissible solution is $a_1=\dfrac{a_0}{\sqrt{a_0^2+2}}\approx 0.871,b_1=c_1=\dfrac{1}{\sqrt{a_0^2+2}}\approx 0.347$; the minimal polynomial of $a_1$ is $p(u)=9u^8-12u^6+18u^4-12u^2+1$ and of $b_1=c_1$ is $q(u)=36u^8-48u^6+36u^4-12u^2+1$. 
Then $\cos(\phi_1)=g(a_1,b_1,c_1)=\dfrac{a_1^3+b_1^3+c_1^3}{\sqrt{a_1^4+b_1^4+c_1^4}}$ and $\phi_1$ are also constructible under the same conditions.

With the essential help of Maple, I solved the problem using the Lagrange method. We seek the critical points of $g^2(a,b,c)$ under the constraints $a^2+b^2+c^2=1,a,b,c\geq 0$.

Case 1. We are not on the edge of the domain. We obtain after eliminating $c$ and the Lagrange's variable:
i) a degenerate curve $\Gamma$: $2a^6+3a^4b^2+2a^3b^3+3a^2b^4+2b^6-3a^4-6a^2b^2-3b^4+3a^2+3b^2-1=0$.
A numerical calculation shows that the minimum on $\Gamma$ is reached in points as $a \approx 0.659306, b\approx 0.315378$ and is $\approx 0.98611>0.9576$.
After eliminating $b$ (resp. $a$), we find the equation in $a$ (resp. $b$):
ii) $p(a)q(a)(3a^2-1)(12a^{12}-24a^{10}+24a^8-24a^6+15a^4-6a^2+1)=0$. 
$a=\pm 1/\sqrt{3}$ implies $a=b=c$ and $\phi=0$. The last factor gives also a maximum ($\cos(\phi)=1$). Finally, we find the solution above ($p(a)=0$ or $q(a)=0$).
Case 2. We are on the edge of the domain. For example $a=0$. After eliminating $c$ and the Lagrange's variable, we obtain (for $b$ or $c$):
$b(b-1)(b+1)(2b^2-1)(b^4-b^2+1)(4b^8-8b^6+12b^4-8b^2+1)=0$.
Only the last factor gives a local minimum: $b_2\approx 0.3992179614,c_2\approx .9168560516$ and $g(0,b_2,c_2)\approx 0.9751727510>0.9576$.