The title pretty much says it all.
Let $\varphi\triangleq\frac{1+\sqrt 5}{2}$ be the golden ratio.
Let $a,b,c$ be some non-negative numbers such that:
- $\frac{1}{1+b}\leq a\leq 1$
- $\frac{c^2}{a+c}\leq b\leq 1$
- $0\leq c\leq 1$
How can we show that $$\frac{1+c}{a+b}\leq \varphi$$ ?
It is worth mentioning that for $a=b=\varphi^{-1}, c=1$, we get $\frac{1+c}{a+b}=\frac{2}{2\varphi^{-1}}=\varphi$.
The motivation for this question comes from the analysis of the price of anarchy for the following game ($A$ denotes the row-player payoff, $B$ denotes the column's):
$A= \left( \begin{array}{ccc} \frac{1}{1+b} & 1 \\ a & \frac{a^2}{a+c} \\ \end{array} \right) $ $B= \left( \begin{array}{ccc} \frac{b}{1+b} & c \\ b & \frac{c^2}{a+c} \\ \end{array} \right) $
This game is a special case of a family of resource-selection games I'm studying.
If we have $\frac{1}{1+b}\leq a\leq 1$ and $\frac{c^2}{a+c}\leq b\leq 1$, then there exists a (pure) Nash equilibrium where the row player plays strategy 2 and the column player plays strategy 1.
This gives us a lower bound on the price of anarchy: $$PoA\geq \frac{1+c}{a+b}$$
Which I would like to bound. Easy enough to use Wolfram or such to get the result but I want an analytic argument.