<s>[INST] 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! 
 [/INST] 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.
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