Curve minimising distance between two points in $ \mathbb{R}^2 $ with a circular obstacle Minimisation functional problem 
$$ \min_{y} \int _ {(-\beta,0)}^A\sqrt{(\dot{y}(x))^2+1} dx \mbox{ such that } x^2+y(x)^2\geq R^2$$ 
Scaling everything, take $R=1, \beta >1$ 
Starting at $(-\beta,0)$ find the path with minimum distance taking you to $A=(x_0,y_0), y_0 \geq 0 \mbox{ wlog }$ without going through the interior of the unit circle. 
Let $ \alpha $ be the angle between the x axis and the line segment joining $ (-\beta,0)$  with $ A $. If $ \sin \alpha \geq 1/ \beta $, then the curve is that line segment. 
You can try the problem with a square, a rhombus any compact set you like. In fact:
Generalisation: find a curve that minimises the distance between two points $A(x_A,y_A)$ and $ B(x_B,y_B) $ that doesn't go through some obstacles. 
The obstacles are open bounded sets $B_i$, with the curve lying in $ \mathbb{R} ^2 \setminus \cup_{i=1}^{\infty} B_i$.
My difficulty: I am not familiar with inequality constraints. Any advice on how to tackle these problems either analytically or computationally would be appreciated.  
 A: Considering the problem
$$
\min_{y} \int _ {(-\beta,0)}^A\left(\sqrt{(\dot{y}(x))^2+1}\right) dx \mbox{ such that } x^2+y^2\ge R^2
$$
we will try to give an informal introduction to the handling of this problem with the help of slack variables $\epsilon(x)$ and lagrange multipliers $\lambda_1(x),\lambda_2(x),\lambda_3(x)$
First, the inequality is converted into an equivalent equality with the contribution of the slack variable
$$
x^2+y^2 \ge R^2 \equiv x^2+y^2-R^2-\epsilon^2(x) = 0
$$
and then the main functional reads
$$
F(x,z,y,y',\epsilon,\epsilon')=\sqrt{z^2+1}+\lambda_1(x^2+y^2-R^2-\epsilon^2)+\lambda_2(2x+2y -2\epsilon \epsilon'z)+\lambda_3(z-y')
$$
Note the inclusion as restriction also the derivative of $x^2+y^2-R^2-\epsilon^2=0$
Now applying the Euler-Lagrange conditions for stationarity we have
$$
F_y-\left(F_{y'}\right)'=2\lambda_1y+2\lambda_2 z+\lambda_3'=0\\
F_z-\left(F_{z'}\right)'=2\lambda_2y+\frac{z}{\sqrt{z^2+1}}+\lambda_3 = 0\\
F_{\epsilon}-\left(F_{\epsilon'}\right)'=2\epsilon(\lambda'_2-\lambda_1)=0
$$
Assuming $\epsilon(x) \ne 0$ we have $\lambda_1=\lambda'_2$ and the equations reduce to
$$
(2\lambda_2 y)'+\lambda'_3=0\Rightarrow 2\lambda_2 y +\lambda_3 = C_1
$$
and then
$$
2\lambda_2 y +\lambda_3 = C_1\\
2\lambda_2y+\frac{y'}{\sqrt{(y')^2+1}}+\lambda_3 = 0
$$
or
$$
C_1 = \frac{y'}{\sqrt{(y')^2+1}}
$$
with solution
$$
y = \pm \frac{C_1 x}{\sqrt{1-C_1^2}}+C_2
$$
Concluding, for $\epsilon(x)\ne 0\;$ the solutions are $y = y^0_i+m_i(x-x_0^i)\;$ for $i\in \{1,2\}$
The solution for $\epsilon(x) = 0\;$ involves the angular circle segment. 
