Non square Matrix as a Tool for Proving an Inequality

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,

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

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.