Let
\[
x = \frac{a}{b - c}, \qquad y = \frac{b}{c - a} \qquad \text{and} \qquad z = \frac{c}{a - b},
\]
where \(a\), \(b\) and \(c\) are distinct real numbers.
Show that
\[
\begin{pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end{pmatrix}
\begin{pmatrix} a \\ b \\ c \end{pmatrix}
=
\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\]
and use this result to deduce that \(yz + zx + xy = -1\).
Hence show that
\[
\frac{a^2}{(b-c)^2} + \frac{b^2}{(c-a)^2} + \frac{c^2}{(a-b)^2} \geqslant 2.
\]
Let
\[
x = \frac{2a}{b+c}, \qquad y = \frac{2b}{c+a} \qquad \text{and} \qquad z = \frac{2c}{a+b},
\]
where \(a\), \(b\) and \(c\) are positive real numbers.
Using a suitable matrix, show that \(xyz + yz + zx + xy = 4\).
Hence show that
\[
(2a + b + c)(a + 2b + c)(a + b + 2c) > 5(b+c)(c+a)(a+b).
\]
Show further that
\[
(2a + b + c)(a + 2b + c)(a + b + 2c) > 7(b+c)(c+a)(a+b).
\]
Solution:
\(\,\) \begin{align*}
&& \begin{pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end{pmatrix}
\begin{pmatrix} a \\ b \\ c \end{pmatrix} &= \begin{pmatrix} a-xb+xc \\ ay+b-yc \\ -za+zb+c \end{pmatrix} \\
&&&= \begin{pmatrix} a-x(b-c) \\ b-y(c-a) \\ c-z(a-b) \end{pmatrix} \\
&&&= \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
\end{align*}
Notice since \(a,b\) and \(c\) are distinct real numbers the vector \(\langle a,b,c \rangle\) cannot be the zero vector, so the determinant of the matrix is zero, ie \(0= 1(1+yz)+x(y-yz)+x(yz+z) = 1 +yz+yx+zx\).
Notice also then that
\begin{align*}
&& \frac{a^2}{(b-c)^2} + \frac{b^2}{(c-a)^2} + \frac{c^2}{(a-b)^2} &= x^2+y^2+z^2 \\
&&&= (x+y+z)^2 - 2(xy+yz+zx) \\
&&&= 2 + (x+y+z)^2 \geq 2
\end{align*}