Prove an inequality in $\mathbb{R}$ Let  $ p,q \in \mathbb{R}, \; \lambda &gt; 0, p \neq q$ (two points).
 For the two points $x_+, x_{-}$ with 
\begin{align*}
x_+&amp;=p+\lambda\cdot (q-p)\\
x_{-}&amp;=p-\lambda \cdot (q-p)
\end{align*}
I have to prove:
$$|\;\!  x_+-q \!\;| &lt; |x_--q|$$
I started with
\begin{align*}
|x_+-q|&amp;=|p+\lambda\cdot(q-p)-q|\\
&amp;=|p-q|+|\lambda\cdot (q-p)|\\
|x_{-}-q|&amp;=|p-\lambda \cdot (q-p)-q|\\
&amp;=|p-q|-|\lambda \cdot (q-p)|
\end{align*}

It's quit understanding, why $|x_+-q| &lt; |x_--q|$ holds, but I'm definitely not sure, if my "proof" is finished at this point. In particular I don't know if $&lt;$ is already prooved.
We have 
\begin{align*}
|x_+ -q|&amp;= |p+\lambda(q-p)-q|\\
&amp;=|p(1-\lambda ) - q( 1-\lambda)|\\
&amp;=|(p-q)\cdot (1-\lambda)|\\
&amp;=|p-q|\cdot |1-\lambda|
\end{align*}
and 
\begin{align*}
|x_- -q|&amp;= |p-\lambda (q-p) -q|\\
&amp;=|p(1+\lambda) - (1+\lambda)q|\\
&amp;=|p-q|\cdot |1+\lambda|
\end{align*}
We know that 
$$ |x_--q|&gt; |x_+-q| \iff |x_--q|-|x_+-q|&gt;0$$
and 
\begin{align*}
|x_--q|-|x_+-q|&amp;=|p-q|\cdot |1+\lambda|-|p-q|\cdot |1-\lambda|\\
&amp;=|p-q|\cdot (|1+\lambda|-|1-\lambda|)
\end{align*}

As $p\neq q$ we have $|p-q|&gt;0$ hence we only need to prove that 
$$|1+\lambda|-|1-\lambda|&gt;0$$
when $0&lt;\lambda \leq 1$ we have
$$|1+\lambda|-|1-\lambda|=1+\lambda - 1 +\lambda = 2 \lambda &gt; 0$$
When $\lambda &gt;1 $ we have 
$$|1+\lambda|-|1-\lambda|=1+\lambda +1-\lambda=2 &gt;0$$
Hence $$|x_--q|&gt;|x_+-q|$$
This result is true for any metric space, whether it's Euclidean or not (or, may be not, according to a comment below).

The initial form of the question, didn't demand the sets to be Euclidean, so is my proof focused on earlier version of the question.
My take was something like this:

$d(x_+,q)=d(p+λ(q−p),q)=d(p,q)+ d(λ(q−p),q) \\ =d(p,q) -d(\lambda (q-p),q)+2d(\lambda (q-p),q) \\ =d(p-\lambda(q-p), q)+ 2d(\lambda (q-p),q) \\ =d(x_-,q) +2d(\lambda (q-p),q)$ 

here,  $x_−=p−λ(q−p)$

(( I used, $d(x\pm z,y) = d(x,y)\pm d(z,y)$))

As $2d(\lambda (q-p),q) \gt 0$  $ \forall p\neq q$ and $\lambda \gt 0$

We have, $d(x_+, q) \gt d(x_-, q) $
$\Box$