At http://www.artofproblemsolving.com/Forum/download/file.php?id=44351 , one finds a short proof by Vasile Cirtoaje of the inequality
$$
\sqrt{8(a^2+bc)+9}+\sqrt{8(b^2+ac)+9}+\sqrt{8(c^2+ab)+9} \geq 15 \ (\text{when} \ a,b,c\gt 0, \ a+b+c=3)
$$
Unfortunately, this proof has several surprising, seemingly un-motivated steps, giving
the impression that only a pure genius could have found them.
So, can anyone “demistify” this proof, possibly transforming some parts into longer but
conceptually more natural ones ?
Related : Prove $\sqrt{x^2+yz+2}+\sqrt{y^2+zx+2}+\sqrt{z^2+xy+2}\ge 6$, given $x+y+z=3$ and $x,y,z\ge0$
I’d rather leave this as a comment, but it’s probably too long so I’ll leave it here.
$$C(a-b)^2 + B(a-c)^2 + A(b-c)^2 \ge 0$$
where $A,B,C$ are polynomials in $a,b,c$. If $A,B,C \ge 0$ all the time then we are of course done, but even in other situations we can usually say something. For example if the inequality is symmetric, one can suppose that $a \ge b \ge c$, then if $A \ge 0$ and $B+C \ge 0$, then
$$C(a-b)^2 + B(a-c)^2 + A(b-c)^2 \ge (B+C)(a-b)^2 + A(b-c)^2 \ge 0$$
There are many variants along this theme. One noteworthy variant is via the Vornicu-Schur lemma, which seeks to write an inequality of the form
$$A(a-b)(a-c) + B(b-c)(b-a) + C(c-a)(c-b) \ge 0$$
and proceed in a similar fashion.
The natural question then is, how to write something into sum of squares? A natural approach is then to pick out $a-b$, $b-c$, $c-a$ terms as much as possible. For example, if there is a $a^2+bc$ term, it is common to write it as $(a-b)(a-c) + a(b+c)$, for the sake of getting $(a-b)(a-c)$ which hopefully gives us a sum of square/Vornicu-Schur like expression.
For the question at hand, to tackle $\sqrt{8(a^2+bc) + 9}$, one can of course write it as $\sqrt{8(a-b)(a-c) + a(b+c) + 9}$, which does not look very helpful with the square root there. One common strategy to remove square root (and product $(a-b), (a-c)$ like term) is to rationalize the denominator. Solely for the purpose of getting sum of squares expression we would consider $\sqrt{A} – (xa + yb + zc)$ as you have mentioned.
How do we pick $x,y,z$? Unfortunately I don’t have a good answer for this. We definitely want $x+y+z=5$ since we expect each $\sqrt{A} – (xa + yb + zc)$ vanish when $a=b=c$. One explanation for the choice $(x,y,z) = (3,1,1)$, is that when we rationalize $\sqrt{A} – (xa + yb + zc)$, we get $A – (xa+yb+zc)^2$, and we want it to be as simple as possible for the sake of easier analysis. $(x,y,z) = (3,1,1)$ immediately kills off the square terms which is a good thing to have.
(Remark: personally, I would rather try $(x,y,z) = (11/5, 7/5, 7/5)$ – one obtains this if one set
$$8(a^2+bc)+9 – (xa+yb+zc)^2 = (9-x^2)(a-b)(a-c) + (1-y^2)(b-a)(b-c) + (1-z^2)(c-a)(c-b)$$
I didn’t check if this would work though.)