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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 $x,y,z\ge0$

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Ewan Delanoy
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  • Which steps are unnatural to you? –  Jun 29 '13 at 17:19
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    @Sanchez Unnatural step 1 : why replace $\Sigma{\sqrt{A}}$ with $\Sigma{\sqrt{A}-(3a+b+c)}$ ? (of course, it all works out in the end, but that’s not a satisfying explanation in my view). It is natural to replace $\Sigma{\sqrt{A}}$ with $\Sigma{\sqrt{A}-(\alpha a+\beta b+\gamma c)}$, but why choose $(\alpha,\beta,\gamma)=(3,1,1)$ ? By the way, $3a+b+c$ is not the same thing as the $3a-b-c$ appearing in $\sqrt{(3a-b-c)^2+q}$, it is a sort of “conjugate”. It would seem more natural to consider $\sqrt{(3a-b-c)^2+q}-(3a-b-c)$ instead of $\sqrt{(3a-b-c)^2+q}-(3a+b+c)$. – Ewan Delanoy Jun 29 '13 at 20:52
  • @Sanchez Unnatural step 2: why replace $\sqrt{A}-(3a+b+c)$ with $\frac{A-(3a+b+c)^2}{\sqrt{A}+(3a+b+c)}$ ? Of course, it all works out in the end, but that’s not a satisfying explanation in my view). – Ewan Delanoy Jun 29 '13 at 20:53

1 Answers1

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I'd rather leave this as a comment, but it's probably too long so I'll leave it here.

  • A basic strategy in tackling inequalities of few variables is to write things into sum of squares, in brute force way. In the case of three variables we would want to write an inequality as

$$C(a-b)^2 + B(a-c)^2 + A(b-c)^2 \ge 0$$ 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.)