An Inequality with Determinants II

Statement

Inequality With Determinants II

Solution 1

Start with subtracting the first column from the other three:

$\displaystyle\begin{align} \Delta &=\left|\begin{array}{cccc} \,1 & 0 & 0 & 0\\ a & b-a & c-a & d-a\\ a^2 & (b-a)(b+a) & (c-a)(c+a) & (d-a)(d+a)\\ \frac{1}{a^2} & \frac{(b-a)(b+a)}{a^2b^2} & \frac{(c-a)(c+a)}{a^2c^2} & \frac{(d-a)(d+a)}{a^2d^2}\end{array}\right|\\ &=\frac{(b-a)(c-a)(d-a)}{a^2} \left|\begin{array}{ccc} \,1 & 1 & 1\\ b+a & c+a & d+a\\ \frac{b+a}{b^2} & \frac{c+a}{c^2} & \frac{d+a}{d^2}\end{array}\right|\\ &=\frac{(b-a)(c-a)(d-a)}{a^2b^2c^2d^2} \left|\begin{array}{ccc} \,1 & 1 & 1\\ b+a & c+a & d+a\\ c^2d^2(b+a) & b^2d^2(c+a) & b^2c^2(d+a)\end{array}\right|\\ &=\frac{(b-a)(c-a)(d-a)}{a^2b^2c^2d^2}\cdot\Delta '; \end{align}$

where $\Delta'\;$ is being evaluated further:

$\displaystyle\begin{align} \Delta' &=\left|\begin{array}{ccc} 1 & 0 & 0\\ b+a & c-b & d-b\\ c^2d^2(b+a) & d^2(c-b)(ab+ac+bc) & c^2(d-b)(ab+ad+bd)\end{array}\right| \end{align}$

It follows that

$\displaystyle\begin{align} |\Delta |&=\frac{|(b-a)(c-a)(d-a)(c-b)(d-b)(d-c)|(abc+abd+acd+bcd)}{a^2b^2c^2d^2}\\ &\lt\frac{abc+abd+acd+bcd}{a^2b^2c^2d^2}\\ &=\frac{1}{abcd}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right) \end{align}$

because $|(b-a)(c-a)(d-a)(c-b)(d-b)(d-c)|\lt 1.$

Solution 2

First of all, the required inequality is equivalent to

$\displaystyle \mathbb{D}=\left|\begin{array}{cccc} \,a^2 & b^2 & c^2 & d^2\\ a^3 & b^3 & c^3 & d^3\\ a^4 & b^4 & c^4 & d^4\\ 1 & 1 & 1 & 1 \end{array}\right|\lt abcd\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right).$

Note that $\displaystyle \mathbb{D}=-\left|\begin{array}{cccc} \,1 & 1 & 1 & 1\\ a^2 & b^2 & c^2 & d^2\\ a^3 & b^3 & c^3 & d^3\\ a^4 & b^4 & c^4 & d^4 \end{array}\right|.\;$ Set $\displaystyle P(x)=\left|\begin{array}{cccc} \,1 & 1 & 1 & 1\\ x^2 & b^2 & c^2 & d^2\\ x^3 & b^3 & c^3 & d^3\\ x^4 & b^4 & c^4 & d^4 \end{array}\right|.\;$

$P(x)\;$ is a polynomial of degree $4\;$ and $P(b)=P(c)=P(d)=0.\;$ Thus

$P(x)=q(b,c,d)(x-b)(x-c)(x-d)(\alpha x+\beta).$

Indeed, $P(a)=q(b,c,d)(a-b)(a-c)(a-d)(\alpha a+\beta).\;$ Now, since $\mathbb{D}\;$ is symmetric in all four variables, it is easy to show that $q(b,c,d)=(b-c)(b-d)(c-d)\;$ so that

(1)

$P(x)=(b-c)(b-d)(c-d)(x-b)(x-c)(x-d)(\alpha x+\beta).$

On the other hand, in the determinant representation of $P(x)\;$ we get

$\displaystyle\begin{align}P(0)&=\left|\begin{array}{cccc} \,1 & 1 & 1 & 1\\ 0 & b^2 & c^2 & d^2\\ 0 & b^3 & c^3 & d^3\\ 0 & b^4 & c^4 & d^4 \end{array}\right|\\ &=b^2c^2d^2\left|\begin{array}{ccc} \,1 & 1 & 1\\ b & c & d\\ b^2 & c^2 & d^2 \end{array}\right|\\ &=b^2c^2d^2(c-b)(d-b)(d-c). \end{align}$

Comparing this to (1) gives $\beta=bcd.\;$ Thus,

$P(x)=(b-c)(b-d)(c-d)(x-b)(x-c)(x-d)(\alpha x+bcd).$

Using the symmetry of $\mathbb{D}\;$ again, we obtain $\alpha=bc+bd+cd,\;$ and, therefore,

$\mathbb{D}=-P(a)=(b-c)(d-b)(d-c)(b-a)(c-a)(d-a)(abc+abd+acd+bcd)$

and the required inequality follows.

Acknowledgment

The inequality from the book Math Power has been posted at the CutTheKnotMath facebook page by Dan Sitaru. Solution 1 is by Leo Giugiuc; Solution 2 is by Héctor Manuel Garduño Castañeda.

 

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