Let $X$ be a normed linear space and $$d(x,y)=\frac {\|x-y\|} {\sqrt {1+\|x\|^{2}}\sqrt {1+\|y\|^{2}}}$$ Is this a metric?
This question arose from the following post where the answer was given in the affirmative when $X$ is an inner product space. I am sorry that I have not made any progress so far. In the post below I asked the question in a comment and someone suggested that I should post this as a separate question.
How to show that the spherical metric satisfies the triangle inequality?
Some additional information: the triangle inequality for $d$ holds for an abstract norm on $X$ iff it holds in $C[0,1]$ with the supremum norm. This may or may not help in answering the question but it makes the question a bit more interesting because we can work with a specific norm. Justification for this claim: $C[0,1]$ is a universal space for the class of separable Banach spaces in the sense any separable Banach space is isometrically isomorphic to a subspace of $C[0,1]$. In particular any three dimensional subspace of $X$ is isometrically isomorphic to a subspace of $C[0,1]$ and the triangle inequality involves only a three dimensional subspace.
Best Answer
Here is a counter-example: Consider $X = \Bbb R^2$ with the $1$-norm $$ \Vert (x_1, x_2) \Vert = |x_1| + |x_2| $$ and the points $$ x = (1, 0) \,, \quad y = (1, 1) \,, \quad z = (0, 1) \, . $$ Then $$ \Vert x \Vert = \Vert z \Vert = 1 \,, \quad \Vert y \Vert = 2 \, , \\ \Vert x - y \Vert = \Vert y - z \Vert = 1 \,, \quad \Vert x - z \Vert = 2 \, , $$ so that $$ d(x, z) = 1 \, , \quad d(x, y) = d(y, z) = \frac{1}{ \sqrt 2 \sqrt 5} $$ and the triangle inequality $$ d(x, z) \le d(x, y) + d(y, z) \iff 1 \le \sqrt \frac 25 $$ does not hold for these points.
More counter-examples can be constructed with the $p$-norm. Using the same points $x,y, z$ leads to the inequality $$ 2^{2/p} (1 + 2^{2/p}) \le 8 \, . $$ which is not satisfied for sufficiently small $p$, e.g. for $p \le \frac 32$.