Csdrhrt

Refer to the problem below (IMO 2009 Shortlist)

Let $a, b, c$ be positive real numbers such that $$frac{1}{a} + frac{1}{b} + frac{1}{c} = a + b+ c$$
Prove that
$$frac{1}{(2a+b+c)^2} + frac{1}{(2a+b+c)^2} + frac{1}{(2a+b+c)^2} leqslant frac{3}{16} $$

In the solution given by the official short list solution book (Pg 16), it states that

Without loss of generality, we choose $$a +b+c = 1$$.
Thus, the problem becomes
$$ frac{1}{(1 + a)^2}
+ frac{1}{(1 + b)^2}
+frac{1}{(1 + b)^2}
leqslant frac{3}{16}(frac{1}{a} + frac{1}{b} + frac{1}{c})$$
Applying Jensen’s inequality to the function $$f(x) = frac{x}{ (1+ x)^2}$$
,which is concave for $0 ≤ x ≤ 2$
and increasing for $0 ≤ x ≤ 1$, we obtain
$$α
frac{a}{(1 + a)^2} + β
frac{b}{(1 + b)^2} + γ
frac{c}{(1 + c)^2} leqslant
(α + β + γ)
frac{A}{(1 + A)^2}$$
, where $A =frac{αa + βb + γc}{α + β + γ}.$
Choosing $α =
frac{1}{a}
, β =frac{1}{b},$ and $γ = frac{1}{c}$
, we can apply the harmonic-arithmetic-mean inequality
$$A =frac{3}{frac{1}{a} + frac{1}{b} + frac{1}{c}}
≤
frac{a + b + c
}{3}
=
frac{1}{3}
< 1$$
Finally we prove:
$$ frac{1}{(1 + a)^2}
+ frac{1}{(1 + b)^2}
+frac{1}{(1 + b)^2}
leqslant ( frac{1}{a} + frac{1}{b} + frac{1}{c}) frac{A}{(1 + A)^2} leqslant ( frac{1}{a} + frac{1}{b} + frac{1}{c})frac{A}{(1 + frac{1}{3})^2} = frac{3}{16}(frac{1}{a} + frac{1}{b} + frac{1}{c}) $$

However, Evan Chen's solution (Pg 4) differs by having $a +b+c=3$ and setting $f(x) = frac{1}{16x} - frac{1}{(x+3)^2}$ and allowing Jensen to resolve the rest.

The questions are as follows;

1. How do you choose before hand which $a + b+ c =$ to choose?




2. Why particularly does the assumption differ between the 2 solutions?




3. How does one know before hand the choosing of $α =
   frac{1}{a}
   , β =frac{1}{b},$ and $γ = frac{1}{c}$?




4. Where and how did $A =frac{αa + βb + γc}{α + β + γ}$ come from?

Any help would be much appreciated.

1. In a homogeneous inequality, this doesn't matter; and $ldots$




2. $ldots$ there is little difference between the two solutions in this respect, because if you take $a, b, c$ from Problem Shortlist with Solutions [it's on p.14 of the PDF you linked to, by the way, rather than p.16] and write $a' = 3a$, $b' = 3b$, $c' = 3c$, then $a', b', c'$ are the $a, b, c$ of Chen's solution, which homogenises the inequality in exactly the same way.




3. Given the idea of applying Jensen's inequality to $x/(1 + x)^2$, the choice of weights $alpha = 1/a$, $beta = 1/b$, $gamma = 1/c$ then gives you the LHS of the desired inequality (in its transformed homogeneous form).




4. The expression $frac{αa + βb + γc}{α + β + γ}$ will naturally appear in any application of Jensen's inequality to a function evaluated at $a, b, c$ with weights $alpha, beta, gamma$.

Another solution.

We need to show that:
$$frac{3}{16}-sum_{cyc}frac{1}{(2a+b+c)^2}geq0$$ or
$$sum_{cyc}left(frac{3}{16a(a+b+c)}-frac{1}{(2a+b+c)^2}right)geq0$$ or
$$sum_{cyc}frac{(3(2a+b+c)^2-16a(a+b+c))bc}{(2a+b+c)^2}geq0$$ or
$$sum_{cyc}frac{(3b^2+3c^2-6a^2-4ab-4ac+6bc)bc}{(2a+b+c)^2}geq0$$ or

$$sum_{cyc}frac{(2a+3b+3c)((c-a)-(a-b))bc}{(2a+b+c)^2}geq0$$ or
$$sum_{cyc}(a-b)left(frac{(2b+3c+3a)ca}{(2b+c+a)^2}-frac{(2a+3b+3c)bc}{(2a+b+c)^2}right)geq0,$$ which is obvious because
$$(a-b)left((2b+3c+3a)a-(2a+3b+3c)bright)geq0$$ and
$$(a-b)((2a+b+c)^2-(2b+a+c)^2)geq0.$$