6. Find the minimum of the expression

$$
\frac{a+b+c}{2}-\frac{[a, b]+[b, c]+[c, a]}{a+b+c}
$$

where the variables $a, b, c$ are any integers greater than $1$ and $[x, y]$ denotes the least common multiple of the numbers $x, y$.

(Tomáš Jurík)
Solution. Considering the symmetry, it suffices to consider triples $(a, b, c)$ where $a \geqq b \geqq c$. For the "smallest" of them $(2,2,2),(3,2,2),(3,3,2),(3,3,3)$, and $(4,2,2)$, the given expression has values $2, 3/2, 17/8, 7/2$, and $11/4$, respectively. If we show that for all other triples $(a, b, c)$ that already satisfy the condition $a+b+c \geqq 9$, the inequality

$$
\frac{a+b+c}{2}-\frac{[a, b]+[b, c]+[c, a]}{a+b+c} \geqq \frac{3}{2}
$$

holds, it will mean that the sought minimum value is equal to $3/2$. We can equivalently rewrite the given inequality as:

$$
\begin{aligned}
(a+b+c)^{2}-2([a, b]+[b, c]+[c, a]) & \geqq 3(a+b+c), \\
a^{2}+b^{2}+c^{2}+2(a b-[a, b])+2(b c-[b, c])+2(c a-[c, a]) & \geqq 3(a+b+c).
\end{aligned}
$$

Since it is clear that $x y \geqq [x, y]$ for any $x, y$, we can ignore the non-negative double terms on the left side of the last inequality and prove the (stronger) inequality

$$
a^{2}+b^{2}+c^{2} \geqq 3(a+b+c).
$$

From the assumption $a+b+c \geqq 9$ and the Cauchy-Schwarz inequality $3\left(a^{2}+b^{2}+c^{2}\right) \geqq (a+b+c)^{2}$, we get

$$
a^{2}+b^{2}+c^{2} \geqq \frac{(a+b+c)^{2}}{3}=3(a+b+c) \cdot \frac{a+b+c}{9} \geqq 3(a+b+c),
$$

and the proof is complete.

Notes. Instead of the Cauchy-Schwarz inequality, we could have rewritten (1) in the form

$$
\left(a-\frac{3}{2}\right)^{2}+\left(b-\frac{3}{2}\right)^{2}+\left(c-\frac{3}{2}\right)^{2} \geqq \frac{27}{4}
$$

and justified this inequality by squaring the obvious inequalities

$$
a-\frac{3}{2} \geqq \frac{5}{2}, \quad b-\frac{3}{2} \geqq \frac{1}{2} \quad \text{and} \quad c-\frac{3}{2} \geqq \frac{1}{2},
$$

since we are already considering only triples where $a \geqq 4, b \geqq c \geqq 2$.

The approach in the solution also leads to the result that for any integers $a, b, c$ greater than 1, the inequality

$$
\frac{a+b+c}{2}-\frac{[a, b]+[b, c]+[c, a]}{a+b+c} \geqq \frac{a+b+c}{6}
$$

holds.

[^0]: 1 Note that we do not exclude the equality $c=0$. A quadruple with a zero element can be reached by the pair of steps described in the following theorem when $b=2$.