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 $ 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.

Let the minimum be $m$.  Then we have the homogeneous symmetric inequality 
$$(a^3+b^3+c^3)^2 \geqslant m^2(a^2+b^2+c^2)(a^4+b^4+c^4)$$
Using $a+b+c = 3u, ab+bc+ca = 3v^2, abc = w^3$, the above reduces to a quadratic in terms of $w^3$ with positive leading term, so the inequality will hold if it holds for  $b=c=1$ and $b=1, c=0  {(\star)}$ .  Thus it is enough to have the following satisfied for $a \in \mathbb R^+$:
$$\frac{a^3+2}{\sqrt{(a^2+2)(a^4+2)}} ,  \frac{a^3+1}{\sqrt{(a^2+1)(a^4+1)}} \geqslant m$$
It turns out the more restrictive condition is the left most, so we find using single variable calculus the minima for $\dfrac{a^3+2}{\sqrt{(a^2+2)(a^4+2)}}$, which is your best $m \approx 0.9576$.

P.S.:  There is a third condition $(\star)$ as the quadratic minimum could be at a stationary point, and while it is not restrictive in this case, it is not as easily tractable as the others and is omitted above.
