# Problem 1 From the 2016 Pan-African Math Olympiad

### Problem

### Solution

We'll first use the AM-GM inequality:

$\begin{align} (x+1)^2+y^2+1 &= (x^2+y^2) +2x+2\\ &\ge 2(xy+x+1). \end{align}$

The other two summands are modified appropriately. Introduce $a,b,c\,$ via $\displaystyle x=\frac{b}{a},\,$ $\displaystyle y=\frac{c}{b},\,$ $\displaystyle z=\frac{a}{c}.\,$ Note that $\displaystyle xy=\frac{c}{a}.\,$ It the follows that

$\begin{align} \sum_{cycl}\frac{1}{(x+1)^2+y^2+1} &\le \sum_{cycl}\frac{1}{2(xy+x+1)}\\ &=\frac{1}{2}\sum_{cycl}\frac{1}{\displaystyle \frac{c}{a}+\frac{b}{a}+1}\\ &=\frac{1}{2}\sum_{cycl}\frac{a}{a+b+c}\\ &=\frac{1}{2}\frac{a+b+c}{a+b+c}=\frac{1}{2}. \end{align}$

### Acknowledgment

Dan Sitaru has kindly posted at the CutTheKnotMath facebook page the above problem from the $24^{th}\,$ Pan-African Mathematical Olympiad, along with his solution. Dan had also remarked that the problem was created in 2006 by Cristinel Mortici from Romania

### Inequalities with the Product of Variables as a Constraint

- Problem 1 From the 2016 Pan-African Math Olympiad
- A Cyclic But Not Symmetric Inequality in Four Variables
- Problem 2, the 36th IMO (1995)
- An Inequality with Two Cyclic Sums
- The Roads We Take

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