How can I prove that, if $x,y,z>0$ and $xyz=1$, then $2(x^2+y^2+z^2)+9\geq 5(x+y+z)$ 
How can I prove that, if $x,y,z>0$ and $xyz=1$, then
$$2(x^2+y^2+z^2)+9\geq 5(x+y+z)$$

I used the famous inequality
$$x^2+y^2+z^2+3\geq 2(x+y+z)$$
I got $$2(x^2+y^2+z^2)+9\geq 4(x+y+z)+3\geq 5(x+y+z)$$
But, the last inequality gives $x+y+z\leq3$ which is not correct.
 A: See the post by @arqady here: https://artofproblemsolving.com/community/c6h1293235p6852442
It's a pretty nice solution, so I'll copy it over here:
We prove the stronger inequality
$$xyz+2(x^2+y^2+z^2)+8\ge 5(x+y+z)$$
WLOG suppose $(x-1)(y-1)\ge 0\implies xy\ge x+y-1\implies xyz\ge xz+yz-z$.
Then we want to show that (by simply plugging in this inequality to the original one)
$$2z^2+(x+y-6)z+2x^2+2y^2-5x-5y+8\geq0$$
Let $x+y=2t$. Using the well-known and easy to prove inequality $2(x^2+y^2)\ge (x+y)^2$, it remains to show that
$$z^2+z(t-3)+2t^2-5t+4\ge 0$$
It suffices to show that
$$(t-3)^2\le 4(2t^2-5t+4)$$
Upon expansion and simplification, this is equivalent to $(t-1)^2\ge 0$, which is true.
A: pqr method:
Let $p = x + y + z,\, q = xy + yz + zx,\, r = xyz = 1$.
It suffices to prove that $2(p^2 - 2q) + 9 \ge 5p$ or
$$2p^2 - 4q + 9 - 5p \ge 0.$$
Using $p^3 - 4pq + 9r \ge 0$ (three degree Schur), we have
$$q \le \frac{p^3 + 9}{4p}.$$
It suffices to prove that
$$2p^2 - 4\cdot \frac{p^3 + 9}{4p} + 9 - 5p \ge 0$$
or
$$(p - 3)(p^2 - 2p + 3)/p \ge 0$$
which is true since $p\ge 3\sqrt[3]{r} = 3$ (AM-GM).
We are done.
Remark: Three degree Schur inequality is
$$a(a - b)(a - c) + b(b - c)(b - a) + c(c - a)(c - b) \ge 0.$$
In pqr language (or substitution), it is $p^3 - 4pq + 9r \ge 0$.
