# Problem 3980 from Crux Mathematicorum

### Problem

Below is Leo Giugiuc's solution to problem 3980 from Crux Mathematicorum. The problem is by S. Viswanathan.

### Proof

With $\displaystyle x=\frac{a+b}{a-b},\; y=\frac{b+c}{b-c},\; z=\frac{c+a}{c-a},\;$ we get

$\displaystyle\begin{cases} (1-x)a+(1+x)b=0\\ (1-y)b+(1+y)c=0\\ (1-z)c+(1+z)a=0 \end{cases}$

which can be written in matrix form

$\left(\begin{array}{ccc} \;1-x & 1+x & 0\\ 0 & 1-y & 1+y\\ 1+z & 0 & 1-z \end{array} \right)\left(\begin{array}\;a\\b\\c\end{array}\right)=\left(\begin{array}\;0\\0\\0\end{array}\right).$

Consider the determinant $D=\left|\begin{array}{ccc}\;1-x & 1+x & 0\\0 & 1-y & 1+y\\1+z & 0 & 1-z\end{array}\right|.\;$ If $D\ne 0,\;$ the system admits only a trivial solution: $a=b=c=0\;$ which is inconsistent with the premises of the problem. Thus, $D=0.\;$ But $D=xy+yz+zx+1,\;$ implying

$xy+yz+zx=-1.$

On the other hand, this is a consequence of the Rearrangement Inequality that, for real numbers $u,v,w,\;$ $\displaystyle uv+vw+wu\le\frac{(u+v+w)^2}{3},\;$ with equality only when $u=v=w.$

Hence, $\displaystyle (x+y+z)xyz\le\frac{(xy+yz+zx)^2}{3}=\frac{1}{3}.\;$ Remains it to prove that the inequality is strict. (Leo Giugiuc credits Professor Radu Gologan with this observation.) By the above remark, equality could only hold if $xy=yz=zx.\;$ Since at least one of $xy,yz,zx\;$ is negative, if there is equality then all three are negative: $xy=yz=zx\lt 0.\;$ But this is impossible because $xy\cdot yz\cdot zx=(xyz)^2\ge 0.\;$ It follows that indeed

$\displaystyle\sum_{cycl}\frac{a+b}{a-b}\prod_{cycl}\frac{a+b}{a-b}\lt\frac{1}{3}.$

### 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 $\;\left(\displaystyle\left(\sum_{i=1}^{n}\frac{x_i^2}{a_i}\right)\cdot\left(\sum_{1\le i\lt j\le n}a_ia_j(x_iy_j-x_jy_i)^2\right)\ge \sum_{i=1}^{n}a_iy_i^2\right)$
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• 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 $a+ac+b\le 2)$
• 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 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 $\left((n + a - 1)(a - 1)^{n-1} \le a^n\right)$