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?

Write
$$abc-b\gcd(a,c)^2-a\gcd(b,c)^2=c\left(a-\frac{\gcd(a,c)^2}{c}\right)\left(b-\frac{\gcd(b,c)^2}{c}\right)-\frac{\gcd(a,c)^2\gcd(b,c)^2}{c}.$$
Since $ac\ge\gcd(a,c)^2$ and $bc\ge\gcd(b,c)^2$, we see that
$$abc-b\gcd(a,c)^2-a\gcd(b,c)^2\ge-\frac{\gcd(a,c)^2\gcd(b,c)^2}{c}.$$
We only need to show that
$$\frac{\gcd(a,c)^2\gcd(b,c)^2}{c}\leq c\gcd(a,b)^2.$$
But this is equivalent to
$$\gcd(a,c)\gcd(b,c)\leq c\gcd(a,b).$$
However, this follows from
$$\gcd(xy,mn)\geq \gcd(x,m)\gcd(y,n)$$
by taking $x=a$, $y=c$, $m=c$, and $n=b$.  From this proof, we can also see that the inequality $$b\gcd(a,c)^2+a\gcd(b,c)^2-c\gcd(a,b)^2\leq abc$$ is an equality iff $a=c$ or $b=c$.
