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
 A: 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.
A: 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 $s$.
A: 
Three views of the surface:  $2(x^4+y^4+z^4)+2xyz+7-5(x^2+y^2+z^2) = a $  from the positive octant for $ a = \frac{6}{5}$ , $ a = \frac{1}{10} $ , and $ a = 0 $ .
A vestige remains at four octants when $ a = \frac{1}{10} $ ( the inequality is  strict). Still,it's interesting to see why , and how symmetry comes into play.
A: 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.
A: 2(x^4+y^4+z^4) + 2xyz + 7 ≥ 5(x^2+y^2+z^2)
We can see easily that the equality holds for for x = y = z = 1 . And as the equation is symmetric hence we can say that x > y > z
Case 1: when x,y,z<1
Then we can easily see that as on the left side as there is 7 so easily the left hand side is greater than the right hand side
Case 2: when x>1 & y,z<1
So the 2(y^4+z^4) + 2xyz + 5 > 5(y^2+z^2) as the int 5 is there. Now for all x
2(x^4) + 2 > 5(x^2) as 2(x^4) > 5(x^2) for all x > (5/2)^(1/2) but if x < (5/2)^(1/2) then +2 is greater than 5(x^2).
Case 3: when x,y>1 & z<1
2(z^4) + 2xyz + 2 > 5(z^2) now for the rest of the part
2(x^4+y^4) + 5 ≥ 5(x^2+y^2) = 2(x^4+y^4) ≥5(x^2+y^2 - 1)
Now using Tchebychef's Inequality on the  (x^2 > y^2 > 1) and (5/2 , 3/2 , 1) we get the following equation
(5/2)x^2+(3/2)y^2 -(1)1 ≥ (x^2+y^2 - 1)(5/2+3/2+1/2)
and 2(x^4+y^4) + 1 ≥(5/2)x^2+(3/2)y^2 ( this can be said as 2(x^4+y^4)≥(5/2)x^2+(3/2)y^2 but when this doesn't satisfy the +1 makes the LHS>RHS) and hence proved.
Case 3: when x,y,z>1
2xyz + 2 ≥ 0 &
2(x^4+y^4+z^4) + 5 ≥ 5(x^2+y^2+z^2) =2(x^4+y^4+z^4) ≥ 5(x^2+y^2+z^2-1) as
by using Tchebychef's Inequality on the  (x^2 > y^2 > z^2 > 1) and (4,3,2,1) we get the following equation
(4x^2+3y^2+2z^2-1)/4≥ 10(x^2+y^2+z^2-1)/16  = (4x^2+3y^2+2z^2-1)≥5/2(x^2+y^2+z^2-1)
2(x^4+y^4+z^4)≥(4x^2+3y^2+2z^2-1)  =  2(x^4+y^4+z^4)+1≥(4x^2+3y^2+2z^2) and hence proved.
A: Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that $f(w^3)\geq0$, where
$$f(w^3)=2(81u^4-108u^2v^2+18v^4+12uw^3)+2w^3+7-5(9u^2-6v^2).$$
But, $f$ is a linear function, which says that it's enough to prove our inequality
for an extremal value of $w^3$.
We know that $x$, $y$ and $z$ are non-negative roots of the equation
$$(X-x)(X-y)(X-z)=0$$ or
$$X^3-3uX^2+3v^2X-w^3=0$$ or
$$w^3=X^3-3uX^2+3v^2X,$$
which says that a line $Y=w^3$ and the graph of $Y=X^3-3uX^2+3v^2X$ 
have three common points and $w^3$ gets an extremal value, when the line $Y=w^3$ 
is a tangent line to the graph $Y=X^3-3uX^2+3v^2X$, 
which happens for equality case of two variables, or maybe $w^3=0$.
Id est, it's enough to check two cases.


*

*$w^3=0$.


Let $z=0$.
We need to prove that
$$2(x^4+y^4)+7\geq5(x^2+y^2)$$ or
$$2x^4-5x^2+3.5+2y^4-5y^2+3.5\geq0,$$
which is true because $5^2-4\cdot2\cdot3.5<0$;


*$y=x$.


We need to prove that
$$4x^4+2(z-5)x^2+2z^4-5z^2+7\geq0,$$
for which it's enough to prove that
$$(z-5)^2-4(2z^4-5z^2+7)\leq0$$ or
$$(z-1)^2(8z^2+16z+3)\geq0.$$
Done!
