# Difficult Inequality [closed]

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

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

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why ? –  Anthony Quas Jul 31 '12 at 1:24
Are you asking me? What I said is simple algebra. –  GH from MO Jul 31 '12 at 1:30
no - i was asking the OP –  Anthony Quas Jul 31 '12 at 2:52
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
@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

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 $$\frac{u}{u+v+d}+\frac{v}{v+u+d}= \frac{u+v}{u+v+d}.$$