Can a Metric Space be Constructed from the Inequality a <= b + c?

  • Thread starter Thread starter pdonovan
  • Start date Start date
  • Tags Tags
    Proof
pdonovan
Messages
17
Reaction score
0

Homework Statement



Show:
if a <= b + c,
then a/(a + 1) <= (b / (b + 1)) + (c / (c + 1))

I'm really not sure where to start from the initial statement a <= b + c. I'm using this in order to show that the metric space e(a,b) = d(a,b)/(1+d(a,b)) exists. Any help would be greatly appreciated, thank you!
 
Physics news on Phys.org
Also, this is for all a,b,c >= 0.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top