Consider all non-negative numbers $a$ and $b$ such that $0 \leq a < b$, and for all $x$, the function

$$
P(x)=a+b \cos x+\frac{b}{2 \sqrt{2}} \cos 2 x
$$

is non-negative.

Determine the set of possible values for the ratio $\frac{b+a}{b-a}$.
Given that $\cos 2 x=2 \cos ^{2} x-1$,

$$
P(x)=\frac{b}{\sqrt{2}}\left(\cos x+\frac{1}{\sqrt{2}}\right)^{2}+a-\frac{b}{\sqrt{2}}
$$

The first term is non-negative and can be 0, for example, when $x=-\frac{\pi}{4}$. Therefore, $P(x)$ will be non-negative for all $x$ if and only if $a-\frac{b}{\sqrt{2}} \geq 0$, which, given $b>0$, is equivalent to the condition

$$
\frac{a}{b} \geq \frac{1}{\sqrt{2}}
$$

Let's denote the value of the ratio $\frac{a}{b}$ by $c$. Since $\frac{b+a}{b-a}=\frac{1+c}{1-c}$ and according to our assumption $\frac{a}{b}<1$, based on (1), we need to examine the value set of the function

$$
\frac{1}{\sqrt{2}} \leq c<1
$$

defined on the interval

$$
Q(c)=\frac{1+c}{1-c}=\frac{2}{1-c}-1
$$

If $c$ runs through the interval (2), $1-c$ runs through the interval

$$
0<1-c \leq 1-\frac{1}{\sqrt{2}}
$$

and its reciprocal runs through the numbers not less than $1 /\left(1-\frac{1}{\sqrt{2}}\right)=2+\sqrt{2}$. Therefore, the value set of $Q(c)$ and thus the set of possible values of the ratio $\frac{b+a}{b-a}$ consists of numbers not less than $(3+2 \sqrt{2})$.