Show that if $x,y\in [0,1]$ and $|x-y|\leq a + b$ then there exists $z\in [0,1]$ such that $|x-z|\leq a$ and $|z-y|\leq b.$ Show that if $x,y\in [0,1], a,b\geq 0$ and $|x-y|\leq a + b$ then there exists $z\in [0,1]$ such that $|x-z|\leq a$ and $|z-y|\leq b.$ 
I tried to come up with a general $z$ that works but I am unable to do so. Any hints would be much appreciated.
 A: If $a,b\in\mathbb{R}$ have no restriction is false because you can choose one of them negative, so you can't find $z$ which satisfy the request.
However, if $a,b\ge 0$ you can reason like this: suppose $x<y$ and take $z=\frac{a}{a+b}y+\frac{b}{a+b}x$.
By choice we have $z\in[x,y]\subseteq[0,1]$, and moreover
\begin{align}
|x-z|=z-x &=\frac{a}{a+b}y+\frac{b}{a+b}x-x=\\
&=\frac{a}{a+b}y+\frac{b-a-b}{a+b}x=\\
&=\frac{a}{a+b}y-\frac{a}{a+b}x=\\
&=\frac{a}{a+b}(y-x)
\end{align}
So $|z-x|=\frac{a}{a+b}|y-x|<\frac{a}{a+b}(a+b)=a$.
In the same way you can prove $|z-y|<b$.
A: Consider the intervals $[x-a,x+a]$ and $[y-b,y+b]$. If they have a point in common then a common point $z$ will do the trick. Prove that neither interval lies completely to the left of the other , i.e. we cannot have $y+b <x-a$ or $x+a<y-b$. This proves that the intervals intersect. Consider the case $x-a \leq y-b\leq x+a \leq y+b$. If the  interval $[y-b,x+a]$ does not intersect $[0,1]$ then either $y-b >1$ or $x+a<0$ and both these are clearly false. Hence we can choose $z$ to be between $0$ and $1$. Similarly argument holds in the other case. 
