Supremum $\sup_x \left\{x-\ln\left(1+\frac{\sin x-\cos x}{2}\right)\right\}$ I would like to know the supremum
$$\sup_x \left\{x-\ln\left(1+\frac{\sin x-\cos x}{2}\right)\right\}$$
when $0\leq x \leq \frac{\pi}{4}$.
After the derivative is easily found that the function is increasing. But, is there a way that does not involve the derivative to find this sup?
Thanks!

Let
$$ f(x)=\frac{e^x}{1+\frac{\sin x-\cos x}{2}}. $$
We only have to show $f(x)$ is increasing in $[0,\frac{\pi}{4}]$. In fract, for $0<x_1<x_1+\alpha<\frac{\pi}{4}$ ($\alpha\in(0,\frac{\pi}{4})$),
\begin{eqnarray*}
\frac{f(x_1+\alpha)}{f(x_1)}&=&e^{\alpha}\frac{2+\sin x_1-\cos x_1}{2+\sin (x_1+\alpha)-\cos (x_1+\alpha)}=e^{\alpha}\frac{\sqrt{2}+\sin(x_1-\frac{\pi}{4})}{\sqrt{2}+\sin(x_1+\alpha-\frac{\pi}{4})}. 
\end{eqnarray*}
It is not hard to check
$$ \sin(x_1+\alpha-\frac{\pi}{4})-\sin(x_1-\frac{\pi}{4})\le \alpha. $$
So
$$\sin(x_1+\alpha-\frac{\pi}{4})\le \alpha+\sin(x_1-\frac{\pi}{4})$$
and hence
\begin{eqnarray*}
\frac{f(x_1+\alpha)}{f(x_1)}\ge e^{\alpha}\frac{\sqrt{2}+\sin(x_1-\frac{\pi}{4})}{\sqrt{2}+\alpha+\sin(x_1-\frac{\pi}{4})}=e^{\alpha}\frac{\sqrt{2}-\sin(\frac{\pi}{4}-x_1)}{\sqrt{2}+\alpha-\sin(\frac{\pi}{4}-x_1)}.
\end{eqnarray*}
Note the following
$$ g(x)=\frac{\sqrt{2}-x}{\sqrt{2}+\alpha-x}, x\in[-1,0] $$
is decreasing. Since $-1<\sin(\frac{\pi}{4}-x_1)\le 0$, we have
\begin{eqnarray*}
\frac{f(x_1+\alpha)}{f(x_1)}&\ge& e^{\alpha}\frac{\sqrt{2}-\sin(\frac{\pi}{4}-x_1)}{\sqrt{2}+\alpha-\sin(\frac{\pi}{4}-x_1)}\\
&=&e^\alpha g(\sin(\frac{\pi}{4}-x_1))\ge e^\alpha g(0)\\
&=&e^\alpha\frac{\sqrt{2}}{\sqrt{2}+\alpha}\\
&\ge&(1+\alpha)\frac{1}{1+\frac{\alpha}{\sqrt{2}}}\ge 1.
\end{eqnarray*}
This implies that $f(x)$ is increasing and so is the original function. So the original function reaches its max at $x=\pi/4$.
