In a metric space $(M,d)$ the triangle inequality $d (x, z) \le d(x, y) + d (y, z)$ gives us's the inequalitie $$ \quad d(x,y)^2 \ge d(x,z)^2 - d(y,z)^2\;\color{}{{-2\cdot d(x,y)\cdot d(y,z)}} $$ which is obtained by raising both sides of the triangle inequality squared.
Question. There are appropriate assumptions under which this inequality can be reversed by replacing the quantity $-2\cdot d(x,y)\cdot d(y,z)$ by another quantity, say $\varphi\big(x,y,z,d(x,z),d(y,z) \big)$ where $\varphi$ is some continuous function $\varphi:M^3\times [0,\infty)^2\to[0,\infty)$ to get a sharp inequality $$ (\ast)\quad d(x,y)^2 \le d(x,z)^2 - d(y,z)^2\color{}{+\varphi\big(x,y,z,d(x,y),d(y,z) \big)}? $$
By appropriate assumptions, I mean since conditions on $ (M, d) $ (e.g. sequential compactness, convexity, uniform convexity, M is a Hadamard manifold, etc) or about $ \varphi $ ( e.g. convexity, compactness, monotonicity, etc) such that inequality $(\ast)$ is true. Trivial examples are $$\varphi\big(x,y,z,d(x,y),d(y,z)\big)=\Big( d(x,y)+d(y,z) \Big)^2 \\ \varphi\big(x,y,z,d(x,y),d(y,z)\big)=\epsilon +2\cdot d(y,z)\cdot d(x,y)+d(y,z)^2,\qquad \epsilon>0.$$
Thanks to everyone.
Update.
I know the question may seem a bit vague, but I understand as valid.The reason for my question is the following. At some time ago looking at the SCOPUS database found an article from 1992 (if I'm wrong about the year will be by a margin of error of no more than 2 years) that dealt with this type inequalities in exhaustive detail. By an unhappiness, I forgot to download the article and do not even remember the title or the author (was a single author). Never found this article again. The incessant searches in various databases that I made on the internet were unsuccessful.
I would be very happy if someone answer me with a link to this article. Incidentally, this would be an answer that I mark as accepted to my question.
As for the article you were looking for I found by doing a search by the top 100 most cited articles in 2013 in MathSciNet.