# An Inequality with Constraint in Four Variables III

### Solution

Denote $ab+ac+ad+bc+bd+cd=q.\,$ The inequality rewrites as $\displaystyle abcd+\frac{15}{2q}\ge\frac{9}{16-2q}.\,$ We know that $abcd\ge q-5.\,$ Consider two cases:

Case 1: $5\le q\le 6$

Suffice it to show that $\displaystyle q-5+\frac{15}{2q}\ge \frac{9}{16-2q}\,$ but this is equivalent to $(q-2)(q-5)(6-q)\ge 0,\,$ which is true in this case.

Case 1: $0\lt q\le 5$

As $abcd\ge 0,\,$ suffice it to show that $\displaystyle\frac{5}{q}\ge\frac{3}{8-q}.\,$ This is so because, in this case, $\displaystyle\frac{5}{q}\ge 1\ge\frac{3}{8-q}.$

### Acknowledgment

The problem and the Solution have been kindly posted by Leo Giugiuc at the CutTheKnotMath facebook page. The problem has been originally posted at the artofproblemsolving forum and is due to Marius Stanean.