Let
\[
y=\dfrac{x^2+x\sin\theta+1}{x^2+x\cos\theta+1}
\,.\]
Given that
\(x\) is real, show that
\[
(y\cos\theta -\sin\theta)^2 \ge 4 (y-1)^2
\,.
\]
Deduce that
\[
y^2+1
\ge
4(y-1)^2
\,,
\]
and hence that
\[
\dfrac {4-\sqrt7}3
\le y \le
\dfrac {4+\sqrt7}3 \,.
\]
In the case $y=
\dfrac {4+\sqrt7}3 \,$, show that \[\sqrt{y^2+1}=2(y-1)\]
and find the corresponding values of \(x\) and
\(\tan\theta\).
Solution:
\(\,\) \begin{align*}
&& y&=\frac{x^2+x\sin\theta+1}{x^2+x\cos\theta+1} \\
\Leftrightarrow && 0 &= x^2(y-1) + x(y \cos \theta - \sin \theta) + y-1 \\
\Leftrightarrow && 0 &\leq \Delta = (y\cos \theta - \sin \theta)^2 - 4(y-1)^2 \\
\Leftrightarrow && (y\cos \theta - \sin \theta)^2 &\geq 4(y-1)^2
\end{align*}
[Assuming that \(y \neq 1\), if \(y = 1\) then the RHS is \(0\) and it is automatically satisfied].
Notice that \((y\cos \theta - \sin \theta)^2 \leq (y^2+1)(\cos^2 \theta + \sin^2 \theta)\) by Cauchy-Schwarz, so \(y^2 + 1 \geq 4(y-1)^2\).
\begin{align*}
&& y^2 + 1 &\geq 4(y-1)^2 \\
\Leftrightarrow && 0 &\geq 3y^2-8y+3 \\
\text{c.v.} && y&= \frac{8 \pm \sqrt{64-4\cdot3 \cdot 3}}{6} \\
&&&= \frac{4 \pm \sqrt{16-9}}{3} = \frac{4 \pm \sqrt{7}}3
\end{align*}
so \(\frac{4-\sqrt{7}}3 \leq y \leq \frac{4+\sqrt7}3\).
If \(y = \frac{4+\sqrt7}3\) then \(y - 1 = \frac{1+\sqrt7}3\) and since \(y^2+1 = 4(y-1)^2\) taking square roots we obtain \(\sqrt{y^2+1} = 2(y-1)\).
Since equality must hold in our C-S identity, we must have \(\langle y, -1 \rangle\) parallel to \( \langle \cos \theta , \sin \theta \rangle\), ie \(\tan \theta = -\frac{3}{4+\sqrt{7}}\) and
\begin{align*}
&& x & = \frac{-(y \cos \theta - \sin \theta) \pm \sqrt{\Delta}}{2(y-1)} \\
&&&= \frac{\pm2(y-1)}{2(y-1)} \\
&&&= \pm1
\end{align*}