IMC 2010 · Day 1 · P5

Consider the symmetric matrix $$A = \begin{pmatrix} 1 & a & b \\ a & 1 & c \\ b & c & 1 \end{pmatrix}.$$ By the constraint we have $\det A \ge 0$ and $\det \begin{pmatrix} 1 & a \\ a & 1 \end{pmatrix}, \det \begin{pmatrix} 1 & b \\ b & 1 \end{pmatrix}, \det \begin{pmatrix} 1 & c \\ c & 1 \end{pmatrix} \ge 0$. Hence $A$ is positive semidefinite, and $A = B^2$ for some symmetric real matrix $B$.

Let the rows of $B$ be $x$, $y$, $z$. Then $|x| = |y| = |z| = 1$, $a = x \cdot y$, $b = y \cdot z$ and $c = z \cdot x$, where $|x|$ and $x \cdot y$ denote the Euclidean norm and scalar product. Denote by $X = \otimes^n x$, $Y = \otimes^n y$, $Z = \otimes^n z$ the $n$th tensor powers, which belong to $\mathbb{R}^{3^n}$. Then $|X| = |Y| = |Z| = 1$, $X \cdot Y = a^n$, $Y \cdot Z = b^n$ and $Z \cdot X = c^n$. So, the matrix $$\begin{pmatrix} 1 & a^n & b^n \\ a^n & 1 & c^n \\ b^n & c^n & 1 \end{pmatrix},$$ being the Gram matrix of three vectors in $\mathbb{R}^{3^n}$, is positive semidefinite, and its determinant, $1 + 2 (abc)^n - a^{2n} - b^{2n} - c^{2n}$ is non-negative.

The constraint can be written as $$\tag{1} (a - bc)^2 \le (1 - b^2)(1 - c^2).$$ By the Cauchy-Schwarz inequality, $$\begin{align*} \bigl( a^{n-1} &+ a^{n-2} bc + \dots + b^{n-1} c^{n-1} \bigr)^2 \le \bigl( |a|^{n-1} + |a|^{n-2} |bc| + \dots + |bc|^{n-1} \bigr)^2 \le \\ &\le \bigl( 1 + |bc| + \dots + |bc|^{n-1} \bigr)^2 \le \bigl( 1 + |b|^2 + \dots + |b|^{2(n-1)} \bigr) \bigl( 1 + |c|^2 + \dots + |c|^{2(n-1)} \bigr) \end{align*}$$ Multiplying by (1), we get $$(a - bc)^2 \bigl( a^{n-1} + a^{n-2} bc + \dots + b^{n-1} c^{n-1} \bigr)^2 \le (1 - b^2) \bigl( 1 + |b|^2 + \dots + |b|^{2(n-1)} \bigr) (1 - c^2) \bigl( 1 + |c|^2 + \dots + |c|^{2(n-1)} \bigr),$$ $$(a^n - b^n c^n)^2 \le (1 - b^n)(1 - c^n),$$ $$1 + 2 (abc)^n \ge a^{2n} + b^{2n} + b^{2n}.$$