Analytic expression for the min value of $g(t):= \sqrt{(t-1)^2 + a^2}+ b|t|$ subject to $|t-1| \le c$ Disclaimer. Not sure this is MO-level but would really appreciate some help with this. Thanks in advance. Moved from SE.
Let $a,b,c \ge 0$, and define a function $g:\mathbb R \to \mathbb R$ by $g(t) := \sqrt{(t-1)^2 + a^2} + b|t|$. It is clear that  $g$ is convex.
Question. What is an analytic formula for the value  $M(a,b,c)$ of $\min_{t \in \mathbb R} g(t)$ subject to $|t-1| \le c $ ?
N.B.: In the special case where $a=0$,  one can easily compute $$ M(0,b,c) = \min_{|t-1| \le c}|t-1| + b|t| =  m(a,b):= \begin{cases}b,&\mbox{ if }b \le 1,\\ 1+(1-c)_+ (b-1),&\mbox{ else.}\end{cases}.$$
An approximation. Also note that for any $a$, one has $g(t) \le a + |t-1| + b|t|$ for all $t \in \mathbb R$, and so $\max(a,M(0, b,c)) \le M(0,b,c) \le  a+M(0,b,c)$. Thus, we deduce that
$$
M(a,b,c) \asymp \max(a,m(b,c)) \asymp a + m(b,c),
$$
where $m (b,c)$ is as given above, and $u  \asymp v $, means that $u$ and $v$ are within  absolute constant multiples of one another.
 A: First, a few simplifications. Note that $g(-t)\ge g(t)$ and $|-t-1|\ge|t-1|$ if $t\ge0$. So, without loss of generality (wlog) $t\ge0$ and
$$g(t)=\sqrt{(t-1)^2 + a^2} + bt. \tag{1}\label{1}$$
Since $g(t)$ is increasing in $t\ge1$, wlog $t\le1$.
Next, change the variables and the constants according to the formulas
$$u=(t-1)^2,\quad A:=a^2,\quad c_2:=\min(1,c^2),$$
so that $u\in[0,1]$ and $t=1-\sqrt u$. So, the problem reduces to minimizing
$$h(u):=\sqrt{u + A} + b(1-\sqrt u)$$
in $u\in[0,c_2]$ given $A\ge0$, $b\ge0$, and $c_2\in[0,1]$.
Note that $h'(u)$ has the same sign for real $u>0$ as $\frac u{A+u}-b^2$, which increases from $-b^2$ to $1-b^2$ as $u$ increases from $0$ to $\infty$. So, the only critical point of $h$ on $(0,\infty)$ is
$$u_*:=A\frac{b^2}{1-b^2}$$
if $0\le b<1$, and no critical points of $h$ on $(0,\infty)$ if $b\ge1$.
Letting now $h_{\min}$ denote the minimum of $h$ on $[0,c_2]$, we see that the desired minimum of $g$ is
$$
h_{\min}=
\begin{cases}
h(u_*)=b+\sqrt{(1-b^2)A}&\text{ if }0\le b<1\ \&\ u_*\le c_2, \\ 
h(c_2)=\sqrt{c_2 + A} + b(1-\sqrt{c_2})&\text{ if }b\ge1\ \text{or}\ (0\le b<1\ \&\ u_*>c_2).  
\end{cases}
$$
This can be simplified:
$$
h_{\min}=
\begin{cases}
b+a\sqrt{1-b^2}&\text{ if }b^2\le\dfrac{c_2}{a^2+c_2}, \\ 
\sqrt{c_2 + a^2} + b(1-\sqrt{c_2})&\text{ otherwise}.  
\end{cases}
$$
