IRL ABooklet

Algebra Difficulty 5.2 AIME, harder Prove it Ireland

Suppose $a, b, c$ are non-zero real or complex numbers, and $x, y, z$ satisfy the following equations.
$bz + cy = a, \ cx + az = b, \ bx + ay = c.$
Prove that
$\frac{3}{4} \le |x|^2 + |y|^2 + |z|^2,$
and give an example to show that this inequality is best possible.
Deduce, or prove otherwise, that in any triangle $ABC$
$\cos^2 A + \cos^2 B + \cos^2 C \ge \frac{3}{4}.$

Solution

$\begin{aligned} |a|^2 &\le (|b|^2 + |c|^2)(|z|^2 + |y|^2) \\ |b|^2 &\le (|c|^2 + |a|^2)(|x|^2 + |z|^2) \\ |c|^2 &\le (|b|^2 + |a|^2)(|x|^2 + |y|^2), \end{aligned}$
whence
$\frac{|a|^2}{|b|^2 + |c|^2} + \frac{|b|^2}{|c|^2 + |a|^2} + \frac{|c|^2}{|a|^2 + |b|^2} \le 2 (|x|^2 + |y|^2 + |z|^2).$
By Nesbitt's inequality, the LHS exceeds $3/2$ . Hence the result.

If $a = b = c = 1$ , then $x = y = z = 1/2$ , in which case $x^2 + y^2 + z^2 = 3/4$ and so the inequality is the best possible. Since in any triangle,
$\begin{aligned} a &= b \cos C + c \cos B, \\ b &= c \cos A + a \cos C, \\ c &= a \cos B + b \cos A, \end{aligned}$
we obtain the consequence.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.