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

Let me give a solution by Mixing Variables method.
We need to prove $f(x,y,z) \geq 0$ for all $x,y,z\geq 0$ where 
$$f(x,y,z)=2(x^4+y^4+z^4)+2xyz+7 - 5(x^2+y^2+z^2).$$ 
If $x>(y+z)^2$ and $y>(x+z)^2$ and $z>(x+y)^2$ then $x+y+z > 2(x^2+y^2+z^2)$, and thus it suffices to prove $2(x^4+y^4+z^4)+7 \geq \frac{5}{2}(x+y+z)$, which is true because one can easily show that $2x^4+\frac{7}{3}> \frac{5}{2}x$ for all $x$.
Now, WLOG, assume that $0\leq z \leq (x+y)^2$. Denote $s=\dfrac{x^2+y^2}{2}$ and $t=xy$, then we have $s\geq t\geq 0$ and $2(s+t)=(x+y)^2\geq z$, and
\begin{align}
f(x,y,z) = g(s,t,z) &= 2(4s^2-2t^2+z^4) +2tz +7 - 5(2s+z^2).
\end{align}
Consider the $t$-dependent term of the above function: $-4t^2+2tz$. We have $-4t^2+2tz - (-4s^2+2sz) = 2(s-t)(2s+2t-z)\geq 0$. Hence, $$g(s,t,z) \geq g(s,s,z).$$ Now we can proceed to prove $g(s,s,z) \geq 0$, which is quite easy (as shown below), but here I would like to highlight the idea of "Mixing Variables". Note that we have proved that
$$f(x,y,z) = g(s,t,z)\geq g(s,s,z) = f\left(\sqrt{\frac{x^2+y^2}{2}},\sqrt{\frac{x^2+y^2}{2}},z\right)$$ 
for $0\leq z \leq (x+y)^2$, which means that we only need to prove the original inequality for the case $x=y$, which is true because
\begin{align}
f(x,x,z) &= 4x^4+2z^4+2x^2z+7-10x^2-5z^2\\  
&= 4(x^2-1)^2+2(z-1)^2(z+1)^2+ (2x^2-2+1-z)(z-1) \\
& \geq 4(x^2-1)^2+2(z-1)^2+ (2x^2-2+1-z)(z-1) \tag{since $(z+1)^2 \geq 1$}\\
& = a^2+2b^2 + (a-b)b \\
& = a^2+b^2 + ab \\
&\geq 0,
\end{align}
where $a=2(x^2-1)$ and $b=z-1$.
Therefore, we have $f(x,y,z)\geq 0$ with equality if and only if $x=y=z=1$.
Remark. The proof for $g(s,s,z)\geq 0$ is exactly the same as the one for $f(x,x,z)\geq 0$ shown above, by replacing $x^2$ by $s$.
