Lipschitz constant for $f(x) = 1/(1+x^2)$ from the definition of Lipschitz Lipschitz constant for $f(x) = 1/(1+x^2)$ from the definition of Lipschitz 
Bit confused on how to do this from the definition instead of using any derivatives
So far I've reduced it to
$$|f(x) - f(y)| \leqslant |(x+y)(x-y)|$$ by simplification and noting that $f(x) \le 1$.

Correct me if wrong:
$|f(x)-f(y)| =|\dfrac{y^2-x^2}{(1+x^2)(1+y^2)}|$
$= \dfrac{|y-x||x+y|}{(1+x^2)(1+y^2)}.$
Note: $|x+y| \le |x|+|y| \le$
$ \sqrt{x^2+1} + \sqrt{y^2+1}\le$
$(1+ x^2) +(1+y^2).$
Hence: 
$|f(x)-f(y)| \lt$
$ |x-y|( \dfrac{1}{1+y^2} +\dfrac{1}{1+x^2}) \le$
$2 |x-y|$.
Note :$ |x| =\sqrt{x^2} \lt \sqrt{x^2+1} \le x^2+1.$
