Problem 787

Solution. Because of the symmetry it suffices to work with $(a, b, c)$, where $a \geqslant b \geqslant c$. For the "least" of them, that is for $(2,2,2),(3,2,2),(3,3,2),(3,3,3)$, and $(4,2,2)$ the expression in question has values $2,3 / 2,17 / 8,7 / 2$, and $11 / 4$ respectively.

We show that $3 / 2$ is the minimal value, namely we show that $(a, b, c)$, which satisfy $a+b+c \geqslant 9$, also fulfill

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

We change the inequality equivalently:

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

Since $x y \geqslant[x, y]$ for any $x, y$, we neglect the non-negative multiples on the left hand side and we prove the (stronger) inequality

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

The assumption $a+b+c \geqslant 9$ and Cauchy inequality $3\left(a^{2}+b^{2}+c^{2}\right) \geqslant(a+b+c)^{2}$ gives

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

which concludes the prove.

[^0]: 1 Let us point out that we do not exclude $c=0$. This is vital, because we end up with such a quadruple if the wizard uses two steps described further for $b=2$.