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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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I am confused, $x$ is defined in terms of $C$ and $C$ is defined in terms of $x$. – dtldarek Apr 16 '12 at 11:12
corrected. Thank you. – user10753 Apr 16 '12 at 11:15
up vote 2 down vote accepted

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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Thank you for your help. Two questions: 1/ What happens if $C \in [m,F]$ ? does the solution work ? 2/ I didn't understand how did you obtain the second inequality (after "This turns to"). Thank you. – user10753 Apr 16 '12 at 13:15
@user10753: If $C \in [m,F]$ the same formula works, because the terms are squared, so the minus doesn't count. You should try it. The second inequality is obtained using the formula $\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$. – Beni Bogosel Apr 16 '12 at 13:18
Perfect. Thank you. Actualy, I need the solution to prove something a little bit stronger: If we denote $S=distance(C,X)$, then $distance(A,C)−distance(A,x) \geq (distance(B,C)−distance(B,x)) - f(S)$ with $f(S)>0$ For example $f(S)= \alpha. S$ with $alpha>0$ a constant. Do you have a hint on how I can use your solution for this ? Thank you in advance for your help. – user10753 Apr 16 '12 at 14:11
@user10753: the inequality you wrote is weaker, not stronger. – Beni Bogosel Apr 16 '12 at 14:21
Looking at the graph of the function $$ x\mapsto\sqrt{x^2+d^2}\qquad(-\infty<x<\infty)$$ (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 $|z|\geq |y|$. – Christian Blatter Apr 16 '12 at 18:43

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