\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
For the first row/component in \(\ihat \), \begin {align*} \begin {pmatrix} 1 & -x & x \end {pmatrix} \begin {pmatrix} a \\ b \\ c \end {pmatrix} & = 1 \cdot a + (-x) \cdot b + x \cdot c \\ & = a + \frac {-ab}{b - c} = \frac {ac}{b - c} \\ & = a + \frac {ac - ab}{b - c} \\ & = a + (-a) \\ & = 0, \end {align*}
and this is similar for the remaining row and components. Hence, we have \[ \begin {pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end {pmatrix} \begin {pmatrix} a \\ b \\ c \end {pmatrix} = \begin {pmatrix} 0 \\ 0 \\ 0 \end {pmatrix} \] as desired.
If the matrix \[ \begin {pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end {pmatrix} \] is invertible, then we must have \[ \begin {pmatrix} a \\ b \\ c \end {pmatrix} = \begin {pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end {pmatrix}^{-1} \begin {pmatrix} 0 \\ 0 \\ 0 \end {pmatrix} = \begin {pmatrix} 0 \\ 0 \\ 0 \end {pmatrix} \] which is impossible, since \(a\), \(b\) and \(c\) are distinct.
Hence, this matrix is not invertible, and it must have a zero-determinant, meaning \begin {align*} 0 & = \det \begin {pmatrix} 1 & -x & x \\ y & 1 & -y \\ -z & z & 1 \end {pmatrix} \\ & = 1 \cdot 1 \cdot 1 + (-x) \cdot (-y) \cdot (-z) + x \cdot y \cdot z - 1 \cdot (-y) \cdot z - (-x) \cdot y \cdot 1 - x \cdot 1 \cdot (-z) \\ & = 1 - xyz + xyz + yz + xy + xz \\ & = xy + yz + zx + 1, \end {align*}
and hence \[ xy + yz + zx = -1. \]
Since \((x + y + z)^2 \geq 0\), we have \begin {align*} 0 & \leq (x + y + z)^2 \\ & = x^2 + y^2 + z^2 + 2 (xy + yz + zx) \\ & = \frac {a^2}{(b - c)^2} + \frac {b^2}{(c - a)^2} + \frac {c^2}{(a - b)^2} + 2 \cdot (-1), \end {align*}
and hence \[ \frac {a^2}{(b - c)^2} + \frac {b^2}{(c - a)^2} + \frac {c^2}{(a - b)^2} \geq 2, \] as desired.
Consider the matrix \[ \begin {pmatrix} -2 & x & x \\ y & -2 & y \\ z & z & -2, \end {pmatrix} \] and for the first row/component in \(\ihat \), \begin {align*} \begin {pmatrix} -2 & x & x \end {pmatrix} \begin {pmatrix} a \\ b \\ c \end {pmatrix} & = (-2) a + bx + cx \\ & = (-2) a + (b + c) x \\ & = (-2) a + 2a \\ & = 0, \end {align*}
and similarly in the second and third rows/components, and hence \[ \begin {pmatrix} -2 & x & x \\ y & -2 & y \\ z & z & -2, \end {pmatrix} \begin {pmatrix} a \\ b \\ c \end {pmatrix} = \begin {pmatrix} 0 \\ 0 \\ 0 \end {pmatrix}. \]
By similar argument as before, this matrix must have a zero determinant as well, and hence \begin {align*} 0 & = \det \begin {pmatrix} -2 & x & x \\ y & -2 & y \\ z & z & -2, \end {pmatrix} \\ & = (-2) (-2) (-2) + x y z + x y z - (-2) y z - x y (-2) - x (-2) z \\ & = -8 + 2xyz + 2xy + 2yz + 2zx, \end {align*}
and hence \[ xyz + xy + yz + zx = 4, \] as desired.
Hence, consider \[ (x + 1) (y + 1) (z + 1) = xyz + xy + yz + zx + x + y + z + 1 = 5 + x + y + z. \]
Since \(a, b, c\) are all positive real numbers, \(x, y, z\) are as well, and hence \(x + y + z > 0\), giving \[ (x + 1) (y + 1) (z + 1) > 5, \] which means \[ \frac {2a + b + c}{b + c} \cdot \frac {a + 2b + c}{a + c} \cdot \frac {a + b + 2c}{a + b} > 5, \] and hence \[ (2a + b + c) (a + 2b + c) (a + b + 2c) > 5 (b + c) (c + a) (a + b) \] as desired.
Furthermore, notice that \begin {align*} x + y + z & = \frac {2a}{b + c} + \frac {2b}{c + a} + \frac {2c}{a + b} \\ & > \frac {2a}{a + b + c} + \frac {2b}{a + b + c} + \frac {2c}{a + b + c} \\ & = \frac {2(a + b + c)}{a + b + c} \\ & = 2. \end {align*}
Hence, \[ (x + 1) (y + 1) (z + 1) > 7, \] which means \[ \frac {2a + b + c}{b + c} \cdot \frac {a + 2b + c}{a + c} \cdot \frac {a + b + 2c}{a + b} > 7, \] and hence \[ (2a + b + c) (a + 2b + c) (a + b + 2c) > 7 (b + c) (c + a) (a + b) \] as desired.