<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] 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!
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