The question is to show that the pressure in the stationary flow of ideal fluid achieves is maximum value on the boundary (and not at an interior point, unless the pressure is constant).
I've come up with:
If the pressure is constant, then, for whatever value $p$ is, that value is, of course, the maximum. So, assume $p\neq$ constant. From Bernouilli's law, we know that $$ \rho\frac{u^2}{2}+p=\text{ constant} $$ Denote the constant to be $C$. Then, we know that the velocity of the flow achieves it's minimum value along the boundary. That is, $$ u_\text{min}=u_\text{boundary} $$ So, for the velocity $u_0$ of any flow not along the boundary, we have $$ u_0>u_\text{min}=u_\text{boundary} $$ Now, $$ \rho\frac{u^2}{2}+p=C \Rightarrow u=\sqrt{\frac{2(C-p)}{\rho}} $$ So, we have $$ u_0>u_\text{min}\\ \sqrt{\frac{2(C-p_0)}{\rho}}>\sqrt{\frac{2(C-p_{u_\text{min}})}{\rho}}\\ \Rightarrow \frac{2(C-p_0)}{\rho}>\frac{2(C-p_{u_\text{min}})}{\rho}\\ \Rightarrow C-p_0>C-p_{u_\text{min}}\\ \Rightarrow p_{u_{\text{min}}}=p_\text{boundary}>p_0 $$ Hence, we have that the pressure along the boundary is strictly greater than the pressure along any other point in the flow when velocity is at its minimum. That is, $p_\text{boundary}=\boxed{p_\text{max}>p_0,}$ where $p_0$ is any point on the interior.
This solution doesn't sit well with me, though. I don't like that I'm making the assumption that $u$ is a min on the boundary. Is there any way around this?