Different proof for the inequality $b \gcd(a,c)^2+ a \gcd(b,c)^2-c \gcd(a,b)^2 \le abc$?

Question

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?

Answer 1

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

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

Answer 2

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 $b=c$.