I would like to submit a recent answer that I gave (and I have deleted) where someone tolds me that I was "total wrong" this is the following :
Begin
Prove that $$\frac{x^2+y^2+z^2}{2}\geq (\alpha\frac{x^4+y^4+z^4}{\beta})^{0.25}$$ With the condition $x+y+z=0$ and $\sqrt{\frac{x^2+y^2+z^2}{2}}\geq \frac{(2\alpha)^{0.25}}{\beta^{0.25}} $
Now the key of this problem is the following Ferrari's identities :
$$(a^2 + 2ac - 2bc - b^2)^4 + (b^2 - 2ab - 2ac - c^2)^4 + (c^2 + 2ab + 2bc - a^2)^4 = 2(a^2 + b^2 + c^2 - ab + ac + bc)^4 . $$
And
$$(a^2 + 2ac - 2bc - b^2)^2 + (b^2 - 2ab - 2ac - c^2)^2 + (c^2 + 2ab + 2bc - a^2)^2 = 2(a^2 + b^2 + c^2 - ab + ac + bc)^2 . $$
It remains to put :
$x=a^2 + 2ac - 2bc - b^2$
$y=b^2 - 2ab - 2ac - c^2$
$z=c^2 + 2ab + 2bc - a^2$
And finally get :
$$(a^2 + b^2 + c^2 - ab + ac + bc)^2\geq \frac{(2\alpha)^{0.25}}{\beta^{0.25}}(a^2 + b^2 + c^2 - ab + ac + bc)$$
Or $$(p)^2\geq \frac{(2\alpha)^{0.25}}{\beta^{0.25}}p$$
Wich is true with the conditions of the beginning !
End
Could someone tells me If I'm in a wrong way or not ?
That's an interesting angle, but honestly a bit hard to follow as posted.
With your notations, as well as $\,p=a^2 + b^2 + c^2 - ab + ac + bc\,$, Ferrari's identities can be written as:
$$ \begin{align} x^2 + y^2+z^2 &= 2p^2 \\ x^4+y^4+z^4 &= 2p^4 \end{align} $$
Then the problem reduces to proving that $\,\sqrt{\dfrac{2p^2}{2}} \ge \sqrt[4]{\dfrac{2\alpha}{\beta}} \implies \dfrac{2p^2}{2} \ge \sqrt[4]{\dfrac{\alpha\cdot 2p^4}{\beta}}\,$ which is of course trivial since both inequalities are equivalent to $\,|p| \ge \sqrt[4]{\dfrac{2\alpha}{\beta}}\,$.
What's missing, however, is that the substitution you used has yet to be justified:
It is not a priori obvious that the system is solvable in $a,b,c$ for all $x,y,z$ such that $x+y+z=0$.
A secondary point is that you must show why the argument works if $\,a,b,c\,$ turn out to be complex.