2006 Romanian NMO Problems/Grade 8/Problem 4

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Problem

Let $a,b,c \in \left[ \frac 12, 1 \right]$. Prove that

$2 \leq \frac{ a+b}{1+c} + \frac{ b+c}{1+a} + \frac{ c+a}{1+b} \leq 3$.

selected by Mircea Lascu

Solution

It is easy to see that the function $f(a,b,c)=\frac{a+b}{c+1}+\frac{b+c}{a+1}+\frac{c+a}{b+1}$ is convex in each of the three variables (since each term is linear or of the form $\frac{p}{x+q}$ for each variable $x$). Thus, its value is maximized at the endpoints. Checking the values of $f$ for all possible values of $a,b,c$ such that $a,b,c\in \{\frac{1}{2},1\}$ yields a maximum of $3$ as desired.

As for the minimum, we have

$2\le \frac{a+b}{c+1}+\frac{b+c}{a+1}+\frac{c+a}{b+1}$

$\Leftrightarrow 5\le \frac{a+b+c+1}{c+1}+\frac{a+b+c+1}{a+1}+\frac{a+b+c+1}{b+1}$

Applying AM-HM to the right hand side yields

$9\left(\frac{a+b+c+3}{a+b+c+1}\right)^{-1}\le\frac{a+b+c+1}{c+1}+\frac{a+b+c+1}{a+1}+\frac{a+b+c+1}{b+1}$

$\Rightarrow 9\left(1-\frac{2}{a+b+c+3}\right)\le\frac{a+b+c+1}{c+1}+\frac{a+b+c+1}{a+1}+\frac{a+b+c+1}{b+1}$

Obviously, $\frac{2}{a+b+c+3}$ is maximized when $a,b,c$ are minimized. That is, when $a=b=c=\frac{1}{2}$. Thus, we have that

$5\le 9\left(1-\frac{2}{a+b+c+3}\right)$

as desired.

See also

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