# Cyclic Inequality in Three Variables by Marian Cucoanes

### Solution 1

\displaystyle \begin{align} \prod_{cycl}(\sqrt{(a+b)(a+c)}-\sqrt{bc})&=\prod_{cycl}\left(\frac{(a+b)(a+c)-bc}{\sqrt{(a+b)(a+c)}+\sqrt{bc}}\right)\\ &\ge \prod_{cycl}\left(\frac{a(a+b+c)}{\displaystyle \frac{2a+b+c}{2}+\frac{b+c}{2}}\right)\\ &=abc. \end{align}

### Solution 2

\displaystyle \begin{align} \prod_{cycl}(\sqrt{(a+b)(a+c)}-\sqrt{bc}) &=\prod_{cycl}\sqrt{bc}\cdot\prod_{cycl}\left(\sqrt{\frac{(a+b)(a+c)}{bc}}-1\right)\\ &=abc\cdot\prod_{cycl}\left(\sqrt{1+\frac{a(a+b+c)}{bc}}-1\right). \end{align}

Noting that $\displaystyle \sqrt{1-x}\ge 1+\frac{x}{2},\,$ whenever $x\ge 0,$

\displaystyle \begin{align} \prod_{cycl}(\sqrt{(a+b)(a+c)-\sqrt{bc}}) &\ge abc\cdot\prod_{cycl}\left(\sqrt{1+\frac{a(a+b+c)}{bc}}-1\right)\\ &\ge abc\prod_{sym}\left(\frac{1}{2}\frac{a(a+b+c)}{bc}\right)\\ &=abc\frac{abc}{8a^2b^2c^2}(a+b+c)^3\\ &=abc\left(\frac{a+b+c}{2(abc)^{1/3}}\right)^3\\ &=abc\left(\frac{3}{2}\right)^3\left(\frac{a+b+c}{3(abc)^{1/3}}\right)^3\\ &=abc\left(\frac{3}{2}\right)^3\\ &\ge abc, \end{align}

where the last step follows from the AM-GM inequality.

### Acknowledgment

I lifted this problem from the mathematical inequalities facebook group. Both because I loved the problem and also the elegant solutions that followed. The problem is by Marian (Gabi Cuc) Cucoaneş.

Diego Alvariz and Ghimisi Dumitrel gave practically the same solution (Solution 1). Solution 2 is by Shyam Subramanian. One of the solutions is based on an erroneous assertion.