<s>[INST] 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
 [/INST] 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.
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