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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.

299792458
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  • I have found another solution. If you want I am ready to post it. – Michael Rozenberg Nov 15 '18 at 20:03
  • Here is a problem for me, since $$(a+b+c)({1\over a}+{1\over b}+{1\over c})\geq 9$$ we have $a+b+c\geq 3$, so I don't understand either solution. – nonuser Nov 15 '18 at 20:19
  • @MichaelRozenberg Can you explain please this phenomena? – nonuser Nov 15 '18 at 20:23
  • @greedoid After homogenization we need to prove that $\sum\limits_{cyc}\frac{1}{(2a+b+c)^2}\leq\frac{3}{16(a+b+c)}\sum\limits_{cyc}\frac{1}{a}$ and we can assume that even $a+b+c=17$, but we saw that the assuming was $a+b+c=1.$ In your example just $\sum\limits_{cyc}\frac{1}{a}\geq9.$ – Michael Rozenberg Nov 15 '18 at 20:35
  • @greedoid See please the first inequality in my previous post. This inequality is equivalent to the given inequality, but it's homogeneous already. – Michael Rozenberg Nov 15 '18 at 20:40
  • I just don't get it. How can be $$1=a+b+c= \sum 1/a \geq 9$$ @MichaelRozenberg – nonuser Nov 15 '18 at 20:47
  • @greedoid After homogenization the condition is not relevant already. In the general, we always use homogenization for to kill the condition. – Michael Rozenberg Nov 15 '18 at 20:51
  • But you can do this homogenization only iff $a+b+c= 1/a+1/b+1/c$ so how can itbe irelevant? @MichaelRozenberg – nonuser Nov 15 '18 at 20:52
  • After homogenization we can assume that $a+b+c=5\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ because after homogenization the inequality does not depend on the substitution $(a,b,c)\rightarrow k(a,b,c)$ for all $k>0$. It says that the condition is not relevant and we can use another condition. $a+b+c=1$ for example. – Michael Rozenberg Nov 15 '18 at 20:57
  • @greedoid Puzzled, I looked it up in Chen's notes. He begins: "First, we want to eliminate the condition. The original problem is equivalent to $$\frac{1}{(2a+b+c)^2} + \frac{1}{(a+2b+c)^2} + \frac{1}{(a+b+2c)^2} \leqslant \frac{3}{16}\cdot\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{a+b+c}.$$ Now the inequality is homogeneous, so we can assume that $a+b+c=3$." $\ldots$ This old dog has just learned a new trick. :) – Calum Gilhooley Nov 15 '18 at 23:02
  • @Michael Rozenberg if you do have another solution you are welcomed to post it. – 299792458 Nov 16 '18 at 01:12
  • Related : https://math.stackexchange.com/questions/591134/let-a-b-c-be-positive-real-number-proof – Arnaud D. Jun 06 '19 at 10:06

2 Answers2

2

  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 $\alpha, \beta, \gamma$.

Calum Gilhooley
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  • But how do you know before hand (as you were solving the problem for the first time) that the specific choice of weights were as stated? Did it had to do with the initial conditions of $\frac{1}{a} + \frac{1}{b} +\frac{1}{c}$? And how does one derive it from here?
    • – 299792458 Nov 16 '18 at 02:12
  • The same goes for $A$. How do you derive it?
    • – 299792458 Nov 16 '18 at 02:14
  • Unaccountably, I seem to have failed to recognise the important qualification beforehand in both parts 1 and 3 of the question - thus really missing the point of your question altogether! I'm sorry. Also, I don't have any light to shed on how one knows when to apply Jensen's inequality. I don't think I'll go so far as to delete my answer, but I think it is probably best ignored! – Calum Gilhooley Nov 16 '18 at 02:25