# An Inequality with Determinants III

### Solution 1

Following the notations in the diagram below:

$OA=a;\;$ $OB=b;\;$ $OC=c;\;$ $AC^2=a^2+c^2;\;$ $BC^2=b^2+c^2;\;$ $AB^2=a^2+b^2\;$ and $OM=x;\;$ $ON=y;\;$ $OP=z;\;$ $MP^2=x^2+z^2;\;$ $NP^2=y^2+z^2;\;$ $MN^2=x^2+y^2.$

One may recollect that

$\displaystyle V^2(OABC)=\frac{1}{288}\left|\begin{array}{ccccc} \,0 & OA^2 & OB^2 & OC^2 & 1\\ OA^2 & 0 & OA^2+OB^2 & OA^2+OC^2 & 1\\ OB^2 & OA^2+OB^2 & 0 & OB^2+OC^2 & 1\\ OC^2 & OC^2+OA^2 & OC^2+OB^2 & 0 & 1\\ 1 & 1 & 1 & 1 & 0\end{array}\right|$

In other words,

$\displaystyle V^2(OABC)=\frac{1}{288}\left|\begin{array}{ccccc} \,0 & a^2 & b^2 & c^2 & 1\\ a^2 & 0 & a^2+b^2 & a^2+c^2 & 1\\ b^2 & a^2+b^2 & 0 & b^2+c^2 & 1\\ c^2 & a^2+c^2 & b^2+c^2 & 0 & 1\\ 1 & 1 & 1 & 1 & 0\end{array}\right|$

Similarly,

$\displaystyle V^2(OMNP)=\frac{1}{288}\left|\begin{array}{ccccc} \,0 & x^2 & y^2 & z^2 & 1\\ x^2 & 0 & x^2+y^2 & x^2+z^2 & 1\\ y^2 & x^2+y^2 & 0 & y^2+z^2 & 1\\ z^2 & x^2+z^2 & y^2+z^2 & 0 & 1\\ 1 & 1 & 1 & 1 & 0\end{array}\right|$

and the fact that, obviously, $V(OMNP)\le V(OABC),\;$ proves the required inequality.

### Solution 2

We'll use column and row transformations:

\displaystyle\begin{align}\left|\begin{array}{ccccc} \,0 & x^2 & y^2 & z^2 & 1\\ x^2 & 0 & x^2+y^2 & x^2+z^2 & 1\\ y^2 & x^2+y^2 & 0 & y^2+z^2 & 1\\ z^2 & x^2+z^2 & y^2+z^2 & 0 & 1\\ 1 & 1 & 1 & 1 & 0\end{array}\right| &= \left|\begin{array}{ccccc} \,0 & x^2 & y^2 & z^2 & 1\\ x^2 & -x^2 & y^2 & z^2 & 1\\ y^2 & x^2 & -y^2 & z^2 & 1\\ z^2 & x^2 & y^2 & -z^2 & 1\\ 1 & 0 & 0 & 0 & 0\end{array}\right|\\ &=\left|\begin{array}{cccc} \,x^2 & y^2 & z^2 & 1\\ -x^2 & y^2 & z^2 & 1\\ x^2 & -y^2 & z^2 & 1\\ x^2 & y^2 & -z^2 & 1\end{array}\right|\\ &=x^2y^2z^2\left|\begin{array}{cccc} \,1 & 1 & 1 & 1\\ -1 & 1 & 1 & 1\\ 1 & -1 & 1 & 1\\ 1 & 1 & -1 & 1\end{array}\right|\\ &=x^2y^2z^2\left|\begin{array}{cccc} \,1 & 1 & 1 & 1\\ -2 & 0 & 0 & 0\\ 0 & -2 & 0 & 0\\ 0 & 0 & -2 & 0\end{array}\right|\\ &=-x^2y^2z^2\left|\begin{array}{ccc} \,-2 & 0 & 0\\ 0 & -2 & 0\\ 0 & 0 & -2\end{array}\right|\\ &=8x^2y^2z^2. \end{align}

Similarly, the determinant in the right-hand side of the required inequality equals $8a^2b^2c^2,\;$ making the inequality obvious.

### Acknowledgment

The inequality from his book Math Accent has been posted at the CutTheKnotMath facebook page by Dan Sitaru. Dan has later communicated privately a solution (Solution 1) and placed a link to this page at the Romanian Mathematical Magazine.