Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose $x,y\geq 0$ and $b,c,d\geq 1$ are integers. Prove or find a counterexample to the following inequality $$ \frac{1}{1+\frac{b+d}{\sqrt{x^{2}+c^{2}}}} + \frac{1}{1+\frac{c+d}{\sqrt{y^{2}+b^{2}}}}\geq \frac{1}{1+\frac{d}{\sqrt{(x+b)^{2}+(y+c)^{2}}}}. $$

share|improve this question
2  
why ? –  Anthony Quas Jul 31 '12 at 1:24
1  
Are you asking me? What I said is simple algebra. –  GH from MO Jul 31 '12 at 1:30
6  
no - i was asking the OP –  Anthony Quas Jul 31 '12 at 2:52
2  
It looks exactly like a typical math. contest inequality and, as such, it is of easy to medium difficulty (that by itself is quite a hint). Since I strongly suspect that some cheating is taking place, I'll not post an answer until I am convinced that I'm wrong here. I'll abstain from voting to close though because I do not have an irrefutable proof of my suspicion either. –  fedja Jul 31 '12 at 11:04
1  
@Anthony Quas: I am familiar with what the OP is asking, but I don't know the answer. The inequality is obtained when considering whether e certain distance function on discrete sets is a metric, i.e., whether it obeys the triangle inequality, for the case p=2. @ fedja: So it does not come from any contest, nor is there any "cheating" whatever that means. –  kodlu Jul 31 '12 at 11:24
show 4 more comments

closed as too localized by Felipe Voloch, Vladimir Dotsenko, Chris Godsil, Suvrit, Daniel Moskovich Aug 1 '12 at 1:00

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

1 Answer

OK, I'll take kodlu's word as a proof this time. :).

Let $u=\sqrt{x^2+c^2}$, $v=\sqrt{y^2+b^2}$, $w=\sqrt{(x+b)^2+(y+c)^2}$. We need to show that $$ \frac{u}{u+b+d}+\frac{v}{v+c+d}\ge \frac{w}{w+d} $$ Note that $w\le u+v$ (triangle inequality on the plane) and that $t\mapsto \frac{t}{t+d}$ is increasing in $t$, so it will suffice to show that $$ \frac{u}{u+b+d}+\frac{v}{v+c+d}\ge \frac{u+v}{u+v+d}. $$ However, $b\le v$ and $c\le u$, so the LHS is at least $$ \frac{u}{u+v+d}+\frac{v}{v+u+d}= \frac{u+v}{u+v+d}. $$

As I said, I wouldn't be surprised to see it on some decent high school math. contest.

share|improve this answer
    
@ fedja: thanks, it is indeed straightforward. –  kodlu Jul 31 '12 at 21:07
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.