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

Here is my solution.
You want to prove that:
$$
2(x^4+y^4+z^4)+2xyz-5(x^2+y^2+z^2)+7\geq0,
$$
for $x,y,z\geq0$. Let's rewrite this inequality in spherical coordinates:
$$
x=r\sin(\theta)\cos(\varphi),\\
y=r\sin(\theta)\sin(\varphi),\\
z=r\cos(\theta),
$$ 
then:
$$
a(\theta,\varphi)r^4+b(\theta,\varphi)r^3-5r^2+7\geq0,
$$
for $r\geq0$, $\theta,\varphi\in[0,\pi/2]$, where
$$
a(\theta,\varphi)=2\{\cos^4(\theta)+\tfrac{1}{4}[3 + \cos(4\varphi)] \sin^4(\theta)\},
$$
and
$$
b(\theta,\varphi)=\tfrac{1}{2}\sin(\theta)\sin(2\theta)\sin(2\varphi).
$$
These functions $a(\theta,\varphi)$ and $b(\theta,\varphi)$ are simply obtained by writing $2(x^4+y^4+z^4)$ and $2xyz$ in spherical coordinates and using standard trigonometric relations such as $\sin(2\alpha)=2\sin(\alpha)\cos(\alpha)$ and $\cos(\alpha)^4+\sin(\alpha)^4=\tfrac{1}{4}[3 + \cos(4\alpha)]$.
Note that in the region $\theta,\varphi\in[0,\pi/2]$, we have $b(\theta,\varphi)\geq0$.
Next, the derivative of the polynomial $p(r)=a(\theta,\varphi)r^4+b(\theta,\varphi)r^3-5r^2+7$ equals $0$ in three points,
$$
0,\frac{-3 b(\theta,\varphi) \pm \sqrt{160 a(\theta,\varphi) + 9 b^2(\theta,\varphi)}}{8 a(\theta,\varphi)}.
$$
By computing the second derivative we immediately obtain that $r=0$ is a maximum. On the other hand, since $b(\theta,\varphi)\geq0$ in $\theta,\varphi\in[0,\pi/2]$, the other possible solution (which has to be a minimum) is
$$
r_{\rm min}(\theta,\varphi)=\frac{-3 b(\theta,\varphi) + \sqrt{160 a(\theta,\varphi) + 9 b^2(\theta,\varphi)}}{8 a(\theta,\varphi)}.
$$
So we have,
$$
a(\theta,\varphi)r^4+b(\theta,\varphi)r^3-5r^2+7\geq a(\theta,\varphi)r_{\rm min}^4(\theta,\varphi)+b(\theta,\varphi)r_{\rm min}^3(\theta,\varphi)-5r_{\rm min}^2(\theta,\varphi)+7.
$$
By plotting $a(\theta,\varphi)r_{\rm min}^4(\theta,\varphi)+b(\theta,\varphi)r_{\rm min}^3(\theta,\varphi)-5r_{\rm min}^2(\theta,\varphi)+7$ in the region $\theta,\varphi\in[0,\pi/2]$ with any standard program of numerical plotting we obtain something like this

Hence, certainly the minimum of $a(\theta,\varphi)r_{\rm min}^4(\theta,\varphi)+b(\theta,\varphi)r_{\rm min}^3(\theta,\varphi)-5r_{\rm min}^2(\theta,\varphi)+7$ in the region $\theta,\varphi\in[0,\pi/2]$ is $0$, which in rectangular coordinates is reached at the point $x=y=z=1$. That proves the required inequality.
