Significance of assumption in competition inequality questions

Question

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.

Answer 1

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 $\alpha, \beta, \gamma$.

Answer 2

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 $$(a-b)\left((2b+3c+3a)a-(2a+3b+3c)b\right)\geq0$$ and $$(a-b)((2a+b+c)^2-(2b+a+c)^2)\geq0.$$