Prove $2(x^4+y^4+z^4)+2xyz+7\ge 5(x^2+y^2+z^2)$ for $x, y, z \ge 0$ 
Let $x,y,z\ge 0$. Show that
$$2(x^4+y^4+z^4)+2xyz+7\ge 5(x^2+y^2+z^2).$$

my idea: let $$x+y+z=p,xy+yz+xz=q,xyz=r$$
since
$$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=p^2-2q$$
and
$$(xy+yz+xz)^2=x^2y^2+y^2z^2+x^2z^2+2xyz(x+y+z)$$
$$\Longrightarrow x^2y^2+y^2z^2+x^2z^2=q^2-2pr$$and
$$x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(x^2y^2+y^2z^2+x^2z^2)=(p^2-2q)^2-2(q^2-2pr)$$
so
$$\Longleftrightarrow 2[(p^2-2q)^2-2(q^2-2pr)]+2r+7\ge 5(p^2-2q)$$
$$\Longleftrightarrow 2p^4-8p^2q+4q^2+8pr-5p^2+10q+2r+7\ge 0$$
then I can't
This link has a similar problem:
see  this
Maybe this problem can use AM-GM inequality,But I can't.Thank you

I've found a solution. Hugly, but still a solution. The procedure suggests that maybe a nice solution is possible by changing variables and using $(x-y), (x-z), (z+y)$.
The idea is that at a min point of the function we are looking for, conditions on the Hessian and the differential force $x=y=z$, so that the problem reduce to an easy case.
Here the (some) details:
1) Check that the minimum of the function  $2(x^4+y^4+z^4)+2xyz+7-5(x^2+y^2+z^2)$ exists and it is atteined for $x,y,z$ strictly positive.
2) At a minimum point, the Jacobian (the differential) must degenerate and the Hessian must be positive definite (or degenerate). The diagonal entries of the Hessian are $24x^2-10$, $24y^2-10$ and $24z^2-10$, so we get 
$$x\geq\sqrt{\frac{5}{12}}
\qquad
y\geq\sqrt{\frac{5}{12}}
\qquad 
z\geq\sqrt{\frac{5}{12}}$$
3) The Jacobian equations are
$$\begin{array}{l}
4x^3+yz=5x\\
4y^3+xz=5y\\
4z^3+xy=5z\end{array}
\qquad\text{that give}
\qquad
\begin{array}{l}
4x^4+xyz=5x^2\\
4y^4+xyz=5y^2\\
4z^4+xyz=5z^2\\
\end{array}$$
4) Subtracting the firts two equations (of the right side) we get
$$4(x^4-y^4)=5(x^2-y^2)$$
that is
$$4(x-y)(x^3+x^2y+xy^2+y^3)=5(x-y)(x+y)$$
if $(x-y)\neq 0$ we can divide and obtain
$$4(x^3+x^2y+xy^2+y^3)=5(x+y)$$
simiarly, by repalcing $x$ with $z$ we get
$$ 4(z^3+z^2y+zy^2+y^3)=5(z+y)$$
by subtracting:
$$4((x^3-z^3)+(x^2-z^2)y+(x-z)y^2)=5(x-z)$$
if $(x-z)\neq 0$ we can divide and obtain
$$4(x^2+xz+z^2+(x+z)y+y^2)=5$$
which is
$$(x+y)^2+(x+z)^2+(z+y)^2=5/2$$
but now, from point 2) we get that the terms $(x+y)^2,(x+z)^2,(z+y)^2$ are all at least $5/3$
whence
$$5/2=(x+y)^2+(x+z)^2+(z+y)^2>5$$
Contradiction. 
Therefore the minpoint is at $x=y=z$ and we reduce to an easy polynomial in one variable.
