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Let $A$, $B$ two points with $distance(A, B)= 2d >0$. Let $m=mid(A, B)$. That is, $distance(A,m)=distance(B,m)=d$.

Define $L$ to be the line that passes through $m$ and which is perpendicular with $[A,B]$.

Let $P$ be the half-plan defined by $L$ and which contains $B$.

Are the following claims true (edited)?

Claim 1: Given any point $C \in P$, given any point $x \in [C, m]$, it holds that: $distance(A, C)-distance(A, x) \geq distance(B, C) - distance(B, x)$

Claim 2: Let $S=distance(C,X)$. Assume $S>0$. Is it true that

$distance(A,C)−distance(A,x)≥(distance(B,C)−distance(B,x)) + f(S)$ with $f(S)>0$. For example $f(S)=α.S$ with alpha>0 a constant.

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    corrected. Thank you.2012-04-16

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My proof is not geometric at all. I suppose there should be a more elegant solution. As you can see in the figure, denote $E,F$ the projections of $A,B$ on the line $Cm$, and notice that $AE=AF=d$ and $mE=mF=a$. Denote $mC=z$ and $mX=y$. Then your inequality is equivalent to $ \sqrt{(z+a)^2+d^2}-\sqrt{(z-a)^2+d^2}\geq \sqrt{(y+a)^2+d^2}-\sqrt{(y-a)^2+d^2} $ This turns to $ \frac{z}{ \sqrt{(z+a)^2+d^2}+\sqrt{(z-a)^2+d^2}}\geq \frac{y}{\sqrt{(y+a)^2+d^2}+\sqrt{(y-a)^2+d^2}}.$

Denote $ f(y)=\frac{y}{\sqrt{(y+a)^2+d^2}+\sqrt{(y-a)^2+d^2}}$ Then it is enough to prove that this function is increasing (then $z \geq y$ finishes the proof). This function is increasing on $[0,\infty)$ because its derivative is positive. Maybe there is a proof without derivatives, but that's the first that came to my mind. enter image description here

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    Looking at the graph of the function x\mapsto\sqrt{x^2+d^2}\qquad(-\infty (a hyperbola) one immediately sees that $ \Bigl|\sqrt{(z+a)^2+d^2}-\sqrt{(z-a)^2+d^2}\Bigr|\geq \Bigl|\sqrt{(y+a)^2+d^2}-\sqrt{(y-a)^2+d^2}\Bigr| $ when $|z|\geq |y|$.2012-04-16