Let, $p,\alpha \in \left (0,\frac{1}{2} \right ]$ with $p \le \alpha$. Define two functions,
$f(p,\alpha)= 1-p-\alpha+2\alpha p$
$g(p,\alpha)=[(1-\alpha)(1-p)+\alpha p]^2+[(1-\alpha)p+\alpha(1-p)]^2$
I wish to show that, $\max \limits_{p,\alpha \in \left (0,\frac{1}{2} \right ] } \frac{f-g}{1-g} \le \frac{1}{2}$
I tried to carry out derivative tests to determine maxima is quite messy. However, I observe that
$g$ is quadratic in $p$ with, $1-g(p,\alpha) \le \frac{1}{2} \: \:\forall \alpha,p$ and $f(p,\alpha) \ge \frac{1}{2}$ and $g \neq 1 \: \:\forall \alpha,p$
Is there any easier trick to argue that maximum is $\le \frac{1}{2}$ through appropriate upper bounding?
Any help would be really appreciated.
Thanks
One way is as the following:
Write $g$ as a function of $f$:
$g=f^2+(1-f)^2$
Then the function becomes
$W=\frac{f-g}{1-g}=\frac{-2f^2+3f-1}{-2f^2+2f}=1-\frac{1}{2f}$
Take the derivative of $W$, with respect to $f$, and you would get
$\frac{\delta W}{\delta f}=\frac{1}{4f^2}$
Since the derivative is always positive, we need to look for the maximum value that $f$ can achieve. In this case, the maximum is achieved, when $\alpha=p=0$, which gives $f=1$.
If $f=1$ is plugged in the formula for $W$, the result would be
$W=\frac{1}{2}$