Different proof for the inequality $b \gcd(a,c)^2+ a \gcd(b,c)^2-c \gcd(a,b)^2 \le abc$? In Encyclopedia of Distances, there is given a distance on natural numbers without zero: $$d(a,b) = \log\left( \frac{ab}{\gcd(a,b)^2}\right)\,.$$ Taking (again for a reference see the Encyclopedia of Distances), the Schoenberg transform of this metric we get the metric: $$D(a,b) = 1-\exp\big(-d(a,b)\big) = 1-\frac{\gcd(a,b)^2}{ab}\,.$$ Since this metric satisfies the triangle inequality, we get: $$D(a,b) \le D(a,c) + D(c,b)\,,$$ and the inequality, after some algebra, $$b \gcd(a,c)^2+ a \gcd(b,c)^2-c \gcd(a,b)^2 \le abc$$ follows for three natural numbers $a,b,c$. My question is, if there is a more direct proof for this?

Let $U$ be the g.c.d. of $a,b$ and $c$.

Let $ RU$ be the g.c.d. of $a$ and $b$, let $SU$ be the g.c.d. of $a$ and $c$, let $TU$ be the g.c.d. of $b$ and $c$.

Then there are natural numbers $A,B$ and $C$, such that $a=RSUA,b=RTUB,c=STUC.$

The inequality to be proved then becomes the rather obvious one

$$SB+TA-RC\le RSTABC.$$