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 $\max(a,M(0, b,c)) \le M(0,b,c) \le a+M(a,b,c)$. Thus, $M(a,b,c) \asymp \max(a,m(b,c))$ 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.

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.

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 $\max_{t \in \mathbb R} g(t)$ 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 $\max(a,M(0, b,c)) \le M(0,b,c) \le a+M(a,b,c)$. Thus, $M(a,b,c) \asymp \max(a,m(b,c))$ 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.

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 $\max(a,M(0, b,c)) \le M(0,b,c) \le a+M(a,b,c)$. Thus, $M(a,b,c) \asymp \max(a,m(b,c))$ 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.

Disclaimer. Not sure this is MO-level but would really appreciate some help with this. Thanks in advance.

Let $a,b,c \ge 0$, and define $f(t) := \sqrt{(t-1)^2 + a^2}$ and $g(t) := f(t) + b|t|$. It is clear that both $f$ and $g$ are convex.

Question. What is an analytic formula for the value $m(a,b,c)$ of $\max_{t \in \mathbb R} g(t)$ subject to $|t-1| \le c $ ?

N.B.: In the special case where $a=0$, one can easily compute $$ m(0,b,c) = \min_{|t-1| \le c}|t-1| + b|t| = \begin{cases}b,&\mbox{ if }b \le 1,\\ 1+(1-c)_+ (b-1),&\mbox{ else.}\end{cases}.$$

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 $\max(a,m(0,b,c)) \le m(a,b,c) \le a + m(0,b,c)$. Thus, $m(a,b,c) \asymp \max(a,m(0,b,c))$ where $m(0,b,c)$ is as given above, and $u \asymp v $, means that $u$ and $v$ are within absolute constant multiples of one another.

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 $\max_{t \in \mathbb R} g(t)$ 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 $\max(a,M(0, b,c)) \le M(0,b,c) \le a+M(a,b,c)$. Thus, $M(a,b,c) \asymp \max(a,m(b,c))$ 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.