How to find a function $\phi(x)$ such that $\sqrt{1+y^2} - \sqrt{1+x^2} \geq \phi(x) (y-x)$ for each $x,y\in \mathbb{R}$ How to find a function $\phi(x)$ such that $\sqrt{1+y^2} - \sqrt{1+x^2} \geq \phi(x) (y-x)$ for each $x,y\in \mathbb{R}$.
Here are some of my ideas:
Also by applying Mean Value theorem, we know that for every $x,y\in \mathbb{R}$ there exists a constant $c$ between $x,y$ such that 
$$\sqrt{1+y^2} - \sqrt{1+x^2} = \frac{c}{\sqrt{1+c^2}}(y-x).$$
I know that $\phi(0) = 0$, by taking $x=0$; $y= \phi(0)$ and $y=-\phi(0)$. Also that $|\phi|$ is bounded above by $1$. 
Could anyone give me a hint on how to continue? Thanks! 
 A: Since $f(z)=\sqrt{1+z^2}$ is a convex function in $C^1(\mathbb{R})$, you are simply looking for the derivative of $f$. Hence:
$$\phi(x) = f'(x)=\frac{x}{\sqrt{1+x^2}}$$
does the job, and no other function can replace it.
A: I) Firstly, consider the more general problem:

Given function $f:\mathbb{R}\to \mathbb{R}$, find function $\phi:\mathbb{R}\to \mathbb{R}$, so that 
  $$\tag{1}\forall x,y\in \mathbb{R}:~~ f(y)-f(x)~\geq~\phi(x)(y-x).$$

Let us deduce that $f$ must necessarily be concave upward. Proof: Consider a convex linear combination $x=\sum_i t_i y_i$, with $t_i\geq 0$ and $\sum_i t_i=1$. Then
$$\sum_i t_i f(y_i) ~\stackrel{(1)}{\geq}~ \sum_i t_i \left\{f(x)+\phi(x)(y_i-x)\right\}~=~f(x)~=~f\left(\sum_i t_i y_i\right)~,\tag{2}$$
cf. Jensen's inequality.
II) Secondly, consider the problem:

Given differentiable function $f:\mathbb{R}\to \mathbb{R}$, find function $\phi:\mathbb{R}\to \mathbb{R}$, so that 
  $$\tag{3}\forall x,y\in \mathbb{R}:~~ f(y)-f(x)~\geq~\phi(x)(y-x).$$

Equivalently:

Given differentiable function $f:\mathbb{R}\to \mathbb{R}$, find function $\phi:\mathbb{R}\to \mathbb{R}$, so that 
  $$ \left[\forall x,y\in \mathbb{R}:~~ y~>~x~\Rightarrow~ \frac{f(y)-f(x)}{y-x}~\geq~\phi(x)\right]
$$$$\tag{4}~\wedge~ \left[\forall x,y\in \mathbb{R}:~~ y~<~x~\Rightarrow~ \frac{f(y)-f(x)}{y-x}~\leq~\phi(x)\right]  .$$

Hint: By letting $y$ approach $x$, we see that the unique possibility in (4) is to choose $\phi=f^{\prime}$.
It is straightforward to check that if $f$ is also assumed to be concave upward, then indeed $\phi=f^{\prime}$ is a solution to (3).
A: Using: $$\sqrt{A}-\sqrt{B} = \frac{A-B}{\sqrt{A}+\sqrt{B}}$$
So $$\sqrt{1+y^2}-\sqrt{1+x^2} = \frac{y^2-x^2}{\sqrt{1+y^2}+\sqrt{1+x^2}} = (y-x)\frac{y+x}{\sqrt{1+y^2}+\sqrt{1+x^2}}$$
So you need a $\phi(x)$ so that for $y>x$, $$\phi(x)\leq \frac{y+x}{\sqrt{1+y^2}+\sqrt{1+x^2}}$$ and for $y<x$
$$\phi(x)\geq\frac{y+x}{\sqrt{1+y^2}+\sqrt{1+x^2}}$$
So $$\sup_{y<x} \frac{y+x}{\sqrt{1+x^2}+\sqrt{1+y^2}}\leq \phi(x)\leq \inf_{y>x} \frac{y+x}{\sqrt{1+x^2}+\sqrt{1+y^2}}$$
Looks like $\phi(x)=\frac{x}{\sqrt{x^2+1}}$ is the only option, since when $y$ gets close to $x$ from either direction, we see that $\phi(x)$ must be both $\leq$ and $\geq$ this function.
That doesn't prove this function works.
It's not hard to prove that $g_x(y)=\frac{x+y}{\sqrt{1+x^2}+\sqrt{1+y^2}}$ is increasing, which would finish the result.
A: For each x, use  mean value theorem, $$\sqrt{1+y^2} - \sqrt{1+x^2} = \frac{c}{\sqrt{1+c^2}}(y-x), \lim_{y\rightarrow x} c(x,y)=x$$ 
When $y>x$, we get $$\phi(x)\leq \frac{c}{\sqrt{1+c^2}}, 
\forall y>x,\text{take limit } \phi(x)\leq \lim_{y\rightarrow x^+}\frac{c}{\sqrt{1+c^2}}=\frac{x}{\sqrt{1+x^2}}$$
When $y<x$, we get $$\phi(x)\geq \frac{c}{\sqrt{1+c^2}}, 
\forall y<x,\text{take limit } \phi(x)\geq \lim_{y\rightarrow x^-}\frac{c}{\sqrt{1+c^2}}=\frac{x}{\sqrt{1+x^2}}$$
So $\phi(x)=\frac{x}{\sqrt{1+x^2}}, x\in \mathbb{R}$.
