Largest constant for an inequality with powers of three variables Let real $(a,b,c)$, with $a+b+c=0$, and let real positive $p$ with $p\ne1$ and $p\ne2$. 
With some constant $C = C(p)$, the following inequality must hold:
$$
(a^2)^p +  (b^2)^p +  (c^2)^p \ge C \cdot (a^{2} + b^{2} + c^{2} )^p
$$
It is understood that $ (a^2)^p$  is nonnegative, likewise for $b$ and $c$.
Find the largest constant $C(p)$ for which the inequality holds.
Note the power mean inequality, without regard of the condition $a+b+c=0$, gives, for $p \ge 1$,  a (too small) constant $\tilde C = (\frac{1}{3})^{p-1}$ so this appears to be the right behavior w.r.t. (large) $p$. 

By homogeneity, let $a^2 + b^2 + c^2 = 1$. Then we need to establish 
$$
(a^2)^p + (b^2)^p + (c^2)^p \ge C(p) 
$$
The two conditions $a^2 + b^2 + c^2 = 1$ and $a+b+c=0$ form a unit circle in $(a,b,c)$-space. Let's parametrize this circle as follows:
$$
a = \frac{-2}{\sqrt{6}}\cos(\phi)\\
b = \frac{2}{\sqrt{6}}\cos(\phi + \pi/3)\\
c = \frac{2}{\sqrt{6}}\cos(\phi - \pi/3)
$$
One can check that the two conditions are obeyed.
Then we have to show 
$$
f(\phi) = \Big(\frac{2}{3}\Big)^p\Big((\cos^2(\phi))^p + (\cos^2(\phi + \pi/3))^p + (\cos^2(\phi - \pi/3))^p\Big) \ge C(p) 
$$
$
f(\phi)$ is periodic in $\phi$ with period length $\pi/3$. Further, note $
f(\phi)= f(-\phi)$ hence there is an extremum at $\phi =0$.
Notice $
f(\phi = 0 ) =f(\phi = \pi/3 ) = \Big(\frac{2}{3}\Big)^p\Big(1 + 2(\cos^2(\pi/3))^p \Big) = \Big(\frac{2}{3}\Big)^p \Big(1 + 2 \Big(\frac{1}{4}\Big)^p  \Big)$
Another extremum in the first period (and hence, the same value  in all other periods) is obtained at $\phi = \pi/6$. This can be seen by putting $\phi = \pi/6 + \alpha$, leading to 
$$
f(\alpha) = \Big(\frac{2}{3}\Big)^p\Big((\cos^2(\pi/6 + \alpha))^p + (\sin^2(\alpha))^p + (\cos^2(\pi/6 - \alpha))^p\Big) 
$$
Hence $f(\alpha) =  f(-\alpha)$ which shows that $\alpha =0$ is an extremum, and 
$$
f(\phi = \pi/6 )  = \Big(\frac{2}{3}\Big)^p\Big(2 (\cos^2(\pi/6))^p\Big) = 2 \Big(\frac{2}{3}\Big)^p\Big(\frac{3}{4}\Big)^p = \Big(\frac{1}{2}\Big)^{p-1}
$$
Now it is easy to see which of the two extrema is the minimum. We have $f(\phi = 0 ) -f(\phi = \pi/6 ) =  \Big(\frac{2}{3}\Big)^p \Big(1 + 2 \Big(\frac{1}{4}\Big)^p -2\Big(\frac{3}{4}\Big)^p \Big) {\ge \atop \le} 0$ for ${{0 \le p \le 1 \; {\rm{and}} \;  {p \ge 2 }}\atop {1 \le p \le 2 }} $, with equality for $p=1$ and $p=2$.
So we have, for $0 \le p \le 1 \; {\rm{and}} \;  {p \ge 2 }$:
$$
f(\phi  )  \ge f(\phi = \pi/6 )  = \Big(\frac{1}{2}\Big)^{p-1} = C(p)
$$
which corresponds to a minimum at $(a,b,c) = (0,-1,1)$ and multiples and permutations.
For ${1 \le p \le 2 }$, we have 
$$
f(\phi  )  \ge f(\phi = 0 )  = \Big(\frac{2}{3}\Big)^p \Big(1 + 2 \Big(\frac{1}{4}\Big)^p  \Big) = C(p)
$$
which corresponds to a minimum at $(a,b,c) = (-2,1,1)$ and multiples and permutations. $\qquad \Box$
