# Hadamard's Determinant Inequalities and Applications I

### Problem

If $A=(a_{ij})\in M_n(\mathbb{R}),$ then

$\displaystyle (\det A)^2\le\prod_{j=1}^n\left(\sum_{i=1}^na_{ij}^2\right).$

(See, e.g., Hadamard's theorem on determinants.)

### Problem 1, Solution

Consider a $3\times 3$ matrix $A=\left(\begin{array}{ccc}a&1&1\\1&b&1\\1&1&c\end{array}\right).$ One can verify that $\det A=2-a-b-c+abc.$ Thus, the required inequality is a direct consequence of Hadamard's First Theorem.

### Problem 2, Solution

Consider a $4\times 4$ matrix $A=\left(\begin{array}{ccc}1&1&1&1\\1&a&0&0\\1&0&b&0\\1&0&0&c\end{array}\right).$ One can verify that $\det A=abc-a-b-c.$ Thus, the required inequality is a direct consequence of Hadamard's First Theorem.

### Acknowledgment

The above is based on an article Application of Hadamard's Theorems to inequalities by Dan Sitaru and Leo Giugiuc that appeared in the Crux Mathematicorum (v 44, n 1, pp 25-27). I am very much indebted to Dan Sitaru for bringing this article to my attention.

### Linear Algebra Tools for Proving Inequalities

$\;\left(\displaystyle\left(\frac{a}{b-c}\right)^2+\left(\frac{b}{c-a}\right)^2+\left(\frac{c}{a-b}\right)^2\ge 2\right)$
• Linear Algebra Tools for Proving Inequalities: Cauchy-Binet Formula $\;\left(\displaystyle\left(\sum_{i=1}^{n}\frac{x_i^2}{a_i}\right)\cdot\left(\sum_{1\le i\lt j\le n}a_ia_j(x_iy_j-x_jy_i)^2\right)\ge \sum_{i=1}^{n}a_iy_i^2\right)$
• An Inequality from Gazeta Matematica, March 2016 (If $a^2+b^2+c^2=3\,$ then $(a+c)(1+b)\le 4)$
• An Inequality from Gazeta Matematica, March 2016 II (If $x^2+y^2+z^2+t^2=1\,$ then $\;(x+z)(y+t)\le 4)$
• An Inequality from Gazeta Matematica, March 2016 III $\;(a^2+b^2+1\ge a+ab+b)$
• An Inequality from Gazeta Matematica, March 2016 IV (If $a^2+b^2+c^2=1\,$ then $a+ac+b\le 2)$
• Problem 3980 from Crux Mathematicorum $\;\left(\displaystyle\sum_{cycl}\frac{a+b}{a-b}\prod_{cycl}\frac{a+b}{a-b}\lt\frac{1}{3}\right)$
• NonSquare Matrix as a Tool for Proving an Inequality $\;\left(2(a + b + c)((a + 2b + 3c) \ge (\sqrt{b(a+b)} + 2\sqrt{c(b+c)} + \sqrt{a(c+a)})^2\right)$
• An Inequality in Parallelogram of Unit Area $\;\left(a^2+b^2+c^2+d^2+ac+bd\ge\sqrt{3}\right)$
• An Inequality from a Vietnamese Problem Book $\;\left(\displaystyle \frac{a^3+2}{b+2c}+\frac{b^3+2}{c+2a}+\frac{c^3+2}{a+2b}\ge 3\right)$
• Hadamard's Determinant Inequalities and Applications II $\left((n + a - 1)(a - 1)^{n-1} \le a^n\right)$