# Leo Giugiuc and Vasile Cirtoaje's Cyclic Inequality

### Solution

WLOG, assume $a\ge b\ge c\ge d,$ which implies $\displaystyle c+d\le\frac{1}{2}.$ Let's prove that

$\displaystyle \sqrt{\frac{c}{1-c}}+\sqrt{\frac{d}{1-d}}\ge \sqrt{\frac{c+d}{1-(c+d)}}.$

Denote $c+d=2s$ and $cd=p,$ such that, in particular, $\displaystyle \frac{1}{4}\ge s\ge p\ge 0.$ Squaring and rearranging, we obtain an equivalent inequality

(*)

$\displaystyle 1\ge\frac{p(1-s)}{(1-2s)\sqrt{1-2s+p^2}}.$

To prove that, observe that from $\displaystyle \frac{1}{4}\ge s\ge p\ge 0,$ $\displaystyle p(1-s)\le s(1-s)\le \frac{3}{16},$ $f(s)=s(1-s)$ is strictly increasing on $\displaystyle \left[0,\frac{1}{4}\right].$ On the other hand, $(1-2s)\sqrt{1-2s+p^2}\ge (1-2s)^{3/2}$ and on $\displaystyle \left[0,\frac{1}{4}\right]$ function $\displaystyle g(s)=(1-2s)^{3/2}$ is strictly decreasing, so that $\displaystyle g(s)\ge\frac{1}{2\sqrt{2}}\gt\frac{3}{16}.$ Thus

$\displaystyle p(1-s)\le\frac{3}{16}\lt (1-2s)\sqrt{1-2s+p^2},$

implying (*). Note that $\displaystyle \sqrt{\frac{c}{1-c}}+\sqrt{\frac{d}{1-d}}\ge \sqrt{\frac{c+d}{1-(c+d)}}$ reduces to equality when $cd=0.$

The bottom line is the required inequality will follow from $\displaystyle \sqrt{\frac{a}{1-a}}+\sqrt{\frac{b}{1-b}}+\sqrt{\frac{c+d}{1-(c+d)}}\ge 2.$ The latter reduces to

$\displaystyle \sqrt{\frac{a}{b+c'}}+\sqrt{\frac{b}{c'+a}}+\sqrt{\frac{c'}{a+b}}\ge 2,$

where $c'=c+d$ so that $a+b+c'=1.$ The latter is exactly Cirtoaje's inequality, proved earlier.

### Acknowledgment

Leo Giugiuc has kindly communicated to me the above statement that he used to solve a problem from a 2017 Saint Petersburg Mathematical Olympiad. The solution is Leo's.