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! 

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 $\phi(x)=\frac{x}{\sqrt{1+x^2}}, x\in \mathbb{R}$.
