I need to prove the following: let p be a metric on X, and f:[0,infinity)->[0,infinity) s.t: 1.f(0)=0. 2.f is monotonically increasing. 3. f satisfy: f((a+b)/2)>=(f(a)+f(b))/2 prove that: f(p(x,y)) is a metric on X, and that that f(p(x,y)) and p(x,y) are equivalent, i.e that there exists reals: b>a>0 s.t a<=f(p(x,y))/p(x,y)<=b. now to prove the first two consitions for metric is quite easy and i did it, but i find it a bit difficult to prove the triangle inequality, i have a feeling that 3's sign should <=, this way we do get the triangle inequality, am i right? and concerning equivalence of metrics, basically if it's f((a+b)/2)<=(f(a)+f(b))/2, then f(p(x,y))/p(x,y)<=2f(p(x,y)/2)/p(x,y)<=....<=2^nf(p(x,y)/2^n)/p(x,y), so for p(x,y) we get the maximum of the ratios, so i think that basically f(p(x,y))/p(x,y)<=f(1), don't know about the left inequality. any hints? as always your help is appreciated.