I am reading a paper where the author states (without proofing) that the equation
$$ {x}^2 \left(1-2 \sqrt{2}\, {x} \right) \sin ^2(\psi )-{y}^2 \left(1-2 \sqrt{2} \,{y} \right)=0 $$
has no positive roots for $y<3\sqrt{2}$ unless $\sin(\psi)>3\sqrt{2}\,y\sqrt{3(1-2\sqrt{2}\,y)}$
Could someone please explain why is this true?
Suppose $y<\frac{1}{2\sqrt{2}}$ is fixed. Then define \begin{align*} f(x)&={x}^2 \left(1-2 \sqrt{2}\, {x} \right) \sin ^2(\psi )-{y}^2 \left(1-2 \sqrt{2} \,{y} \right)\\ f'(x)&=\left(2x -6 \sqrt{2}\, x^2 \right) \sin ^2(\psi ) \end{align*} Since $f$ is concave down the maximum value occurs at $x_0=\frac{1}{3 \sqrt{2}}$. Plugging it in $f(x)$ gives us \begin{align*} f(x_0)&= \frac{\sin ^2(\psi )}{54}-{y}^2 \left(1-2 \sqrt{2} \,{y} \right) \end{align*}
If $\sin(\psi)<3\sqrt{2}\,y\sqrt{3(1-2\sqrt{2}\,y)}$ then the maximum value of this function is negative hence there is no roots.