# Non square Matrix as a Tool for Proving an Inequality

### Proof

Define matrix

$\displaystyle A=\left(\begin{array}{cccc}\;\sqrt{a+b} & \sqrt{b+c} & \sqrt{a} & \sqrt{c}\\\sqrt{b} & \sqrt{c} & \sqrt{a+c} & \sqrt{b+c}\end{array}\right).\;$ We have $A\in M_{4,2}(\mathbb{R}).\;$ Further

$AA^t=\left(\begin{array}{cc}\;a+b+b+c+a+c & \sqrt{a(a+b)}+2\sqrt{c(b+c)}+\sqrt{a(a+c)}\\ \sqrt{a(a+b)}+2\sqrt{c(b+c)}+\sqrt{a(a+c)} & a+b+b+c+a+c\end{array}\right).$

$AA^t\in M_2(\mathbb{R}).\;$ By the Cauchy-Binet theorem, $\det (AA^t)\ge 0.\;$ More explicitly,

$AA^t=\left(\begin{array}{cc}\;2a+2b+2c & \sqrt{a(a+b)}+2\sqrt{c(b+c)}+\sqrt{a(a+c)}\\ \sqrt{a(a+b)}+2\sqrt{c(b+c)}+\sqrt{a(a+c)} & 2a+2b+2c\end{array}\right).$

whereas,

$\det(AA^t)=(2a+2b+2c)(a+2b+3c)- (\sqrt{b(a+b)} + 2\sqrt{c(b+c)} + \sqrt{a(c+a)})^2.$ Or, else,

$\det(AA^t)=2(a + b + c)((a + 2b + 3c) - (\sqrt{b(a+b)} + 2\sqrt{c(b+c)} + \sqrt{a(c+a)})^2\ge 0.$

### Acknowledgment

The problem and the proof ave been comunicted to me by Dan Sitaru. Both are due to Daniel Sitaru, Leonard Giugiuc, Dr. Tr. Severin, Romania.

### 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 $\;\left(\displaystyle\sum_{cycl}\frac{a+b}{a-b}\prod_{cycl}\frac{a+b}{a-b}\lt\frac{1}{3}\right)$
• 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 I $\left((2-a-b-c+abc)^2\le (a^2+2)(b^2+2)(c^2+2)\right)$
• Hadamard's Determinant Inequalities and Applications II $\left((n + a - 1)(a - 1)^{n-1} \le a^n\right)$