# An Inequality with Determinants VII

### Statement

### Solution

Following the notations in the diagram below:

$OA=a;\;$ $OB=b;\;$ $OC=c;\;$ $AC^2=a^2+c^2;\;$ $BC^2=b^2+c^2;\;$ $AB^2=a^2+b^2\;$

Let

$\displaystyle V=\left|\begin{array}{ccccc} \,0 & OA^2 & OB^2 & OC^2 & 1\\ OA^2 & 0 & OA^2+OB^2 & OA^2+OC^2 & 1\\ OB^2 & OA^2+OB^2 & 0 & OB^2+OC^2 & 1\\ OC^2 & OC^2+OA^2 & OC^2+OB^2 & 0 & 1\\ 1 & 1 & 1 & 1 & 0\end{array}\right|$

One may recollect the von *Staudt-Crelle formula*: $\displaystyle V^2(OABC)=\frac{1}{288}V .$ It was already used on another occasion, where it was also observed that $\displaystyle V(OABC)=\frac{1}{6}abc.$ Now, the required inequality is getting proved with an application of the AM-GM inequality:

$\displaystyle\begin{align} V(OABC)&=\frac{1}{6}abc=\frac{1}{6}\sqrt{ab}\cdot\sqrt{bc}\cdot\sqrt{ca}\\ &\le\frac{1}{6}\cdot\frac{a+b}{2}\cdot\frac{b+c}{2}\cdot\frac{c+a}{2}\\ &=\frac{1}{48}\prod_{cycl}(a+b). \end{align}$

So that

$\displaystyle V=288\cdot V^2(OABC)\le\frac{288}{48^2}\prod_{cycl}(a+b)^2=\frac{1}{8}\prod_{cycl}(a+b)^2.$

### Acknowledgment

The inequality is a follow-up on an earlier one from Dan Sitaru's book *Math Accent*. Dan has kindly communicated to me the problem and the solution on a LaTeX file.

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