Proving radical inequality by a substitution and $pqr$ 
(Zuyong) If $a,b,c>0$ and $abc=1$, prove that
\[a+b+c+6\ge\sqrt{a^2+8}+\sqrt{b^2+8}+\sqrt{c^2+8}.\]

I want to see if teomehai's substitution works, since it does deal with the square roots very well.
We have $a=\dfrac{2-x^2}x$, $b=\dfrac{2-y^2}y$, $c=\dfrac{2-z^2}z$, then $abc=1$ implies that $\prod\left(2-x^2\right)=xyz$, and need to prove that $x+y+z\le\dfrac62=3$.
Let $p=x+y+z$, $q=xy+yz+zx$, $r=xyz$. The condition turns into
$$r^2+r+4p^2+4pr-8-8q-2q^2=0.$$
By Schur inequality, $r\ge\dfrac{4pq-p^3}9$, plug in to get
\begin{align*}&16 {{p}^{2}}{{q}^{2}}-162 {{q}^{2}}-8 {{p}^{4}} q+144 {{p}^{2}} q+36 p q
\\[3pt]{}-{}&648 q+{{p}^{6}}-36 {{p}^{4}}-9 {{p}^{3}}+324 {{p}^{2}}-648\le0\vphantom\strut.\end{align*}
Since $p^6$ can blow up faster than anything else, $p$ might get the desired up bound although I don't know how.

Remark: Here is a complicated proof using pqr method.
Let $2x = \sqrt{a^2 + 8} - a,
2y = \sqrt{b^2 + 8} - b, 2z = \sqrt{c^2 + 8} - c$
(correspondingly, $a=\frac{2-x^2}x$, $b=\frac{2-y^2}y$, $c=\frac{2-z^2}z$).
Let $p = x + y + z, q = xy + yz + zx, r = xyz$.
The condition $abc = 1$ is written as
$$r^2 + r + 4p^2 + 4pr - 8 - 8q - 2q^2 = 0. \tag{1}$$
Also, clearly, $x, y, z \le \sqrt 2$. We have $x + y + z \le 3\sqrt 2$ and
$$(\sqrt 2 - x)(\sqrt 2 - y) + (\sqrt 2 - y)(\sqrt 2 - z)
+ (\sqrt 2 - z)(\sqrt 2 - x) \ge 0$$
which are respectively written as
$$p \le 3\sqrt 2, \tag{2}$$
and
$$6 - 2\sqrt 2\, p + q \ge 0. \tag{3}$$
We need to prove that $x + y + z \le 3$.
We split into two cases:
Case 1: $p^2 \ge 4q$
Using $q \le p^2/4$ and (3), we have
$$6 - 2\sqrt 2\, p + p^2/4 \ge 0$$
or
$$\frac14(6\sqrt 2 - p)(2\sqrt 2 - p) \ge 0$$
which, when combined with (2), results in $p \le 2\sqrt 2 < 3$.
Case 2: $p^2 < 4q$
Using $r \ge \frac{(p^2 - q)(4q - p^2)}{6p} \ge 0$ (degree four Schur), from (1), we have
\begin{align*}
 &16\,{q}^{4}-40\,{p}^{2}{q}^{3}+ \left( 33\,{p}^{4}-168\,{p}^{2}-24\,p
 \right) {q}^{2}\\
 &\qquad + \left( -10\,{p}^{6}+120\,{p}^{4}+30\,{p}^{3}-288\,{p
 }^{2} \right) q+{p}^{8}-24\,{p}^{6}-6\,{p}^{5}+144\,{p}^{4}-288\,{p}^{
  2}\\
 &\le 0. \tag{4}
\end{align*}
Denote LHS by $f(p, q)$.
We can prove that
$$\left. \begin{array}{r}
 6 - \frac{14}{5} p + q > 0\\
 p^2 \ge 3q\\
 p\le 9/2\\
 p > 3
\end{array}
\right\} \Longrightarrow f(p, q) > 0. \tag{5}$$
(The proof is given at the end.)
Note that $6 - 2\sqrt 2 \, p + q \ge 0$ implies $6 - \frac{14}{5}p + q > 0$, and $p \le 3\sqrt 2$ implies $p \le 9/2$. (Note: We avoid irrational numbers.)
From (4) and (5), we get a contradiction. As a result, we have $p \le 3$.
We are done.

Proof of (5):
We have
$$3 < p \le \frac92 
\quad \iff \quad p = 3 \cdot \frac{s}{1 + s} + \frac92 \cdot \frac{1}{1 + s}, \quad s \ge 0. $$
Then we have
$$\frac{14}{5}p - 6 < q \le \frac{p^2}{3} \quad \iff
\quad q = \left(\frac{14}{5}p - 6\right)\cdot \frac{t}{1+t}
+ \frac{p^2}{3}\cdot \frac{1}{1+t}, \quad t \ge 0.$$
We have
$$f(p, q) = \frac{81}{160000(1+s)^8(1+t)^4}[g(s,t) + 4882500] > 0$$
where $g(s,t)$ is a polynomial with non-negative coefficients.
We are done.
