# An Inequality with Determinants IV

### Solution

We shall compute the determinant first using row and/or column operations:

\displaystyle\begin{align} D&=\left|\begin{array}{cccc} \,a & -b & 0 & 0\\ 0 & b & -c & 0\\ 0 & 0 & c & -d\\ 1 & 1 & 1 & 1+d\end{array}\right|\\ &=a\left|\begin{array}{ccc}b&-c&0\\0&c&-d\\1&1&1+d\end{array}\right|+b\left|\begin{array}{ccc}0&-c&0\\0&c&-d\\1&1&1+d\end{array}\right|\\ &=ab\left|\begin{array}{cc}c&-d\\1&1+d\end{array}\right|+a\left|\begin{array}{cc}-c&0\\c&-d\end{array}\right|+b\left|\begin{array}{cc}-c&0\\c&-d\end{array}\right|\\ &=ab(c+cd+d)+acd+bcd\\ &=abcd+abc+abd+acd+bcd\\ &=\frac{1}{2}abcd+\frac{1}{2}abcd+abc+abd+acd+bcd\\ &\ge 6\left(\frac{1}{4}a^5b^5c^5d^5\right)^{1/6}\\ &=3\sqrt[3]{4}(abcd)^{5/6}. \end{align}

The equality holds when

$\displaystyle \frac{1}{2}abcd=abc=abd=acd=bcd,$

i.e., when $a=b=c=d=2.$

### Extra pondering

It was "natural" to apply the AM-GM inequality above to obtain

$abcd+abc+abd+acd+bcd\ge 5(a^4b^4c^4d^4)^{1/5}$

that would lead to $D\ge 5(abcd)^{4/5},\;$ with the equality for $a=b=c=d=1.\;$ Following up on the solution we may ask about inequalities involving the above determinant that become equalities for $a=b=c=d=n,\;$ for $n\;$ other than $1\;$ or $2.\;$ Write

\displaystyle\begin{align} D&=abcd+abc+abd+acd+bcd\\ &=n\frac{1}{n}abcd+abc+abd+acd+bcd\\ &\ge (n+4)\left(n^{-n}a^{n+3}b^{n+3}c^{n+3}d^{n+3}\right)^{1/(n+4)}\\ &=(n+4)n^{-n/(n+4)}(abcd)^{(n+3)/(n+4)}, \end{align}

with the equality for $a=b=c=d=n.$

### Acknowledgment

The inequality has been shared at the CutTheKnotMath facebook page by Dan Sitaru.