Prove Geometry Inequality: 60° ≤ ($aA$+$bB$+$cC$)/($a$+$b$+$c$) < 90°

[magneto1]

(BMO, 2013) The angles $A$, $B$, $C$ of a triangle are measures in degrees, and the lengths of the opposite sides are
$a$,$b$,$c$ respectively. Prove:
\[
60^\circ \leq \frac{aA + bB + cC}{a + b + c} < 90^\circ.
\]

Edit: Update to include the degree symbol for clarification. Thanks, anemone.

 

[Opalg]

[sp]Let $p = a+b+c$ be the perimeter of the triangle. Then $\frac{aA + bB + cC}{a + b + c} = \frac apA + \frac bpB + \frac cpC$ is a weighted mean of the three angles. Each side of a triangle is always less than half the perimeter, so $\frac ap,\, \frac bp,\, \frac cp$ are all less than $\frac12.$ Thus $\frac apA + \frac bpB + \frac cpc < \frac12(A+B+C) = \frac12(180^\circ) = 90^\circ.$

For the other inequality, the largest angle of a triangle is always opposite the longest side, and the smallest angle is opposite the shortest side. (That is "geometrically obvious", but I don't offhand know a proof of it.) So the weighted mean $\frac apA + \frac bpB + \frac cpC$ gives the greatest weight to the largest angle and the least weight to the smallest angle, and therefore must be greater than (or equal to) the unweighted mean $\frac13 A + \frac13 B + \frac13 C = \frac13(180^\circ) = 60^\circ.$[/sp]

 

[magneto1]

[sp]Let $p = a+b+c$ be the perimeter of the triangle. Then $\frac{aA + bB + cC}{a + b + c} = \frac apA + \frac bpB + \frac cpC$ is a weighted mean of the three angles. Each side of a triangle is always less than half the perimeter, so $\frac ap,\, \frac bp,\, \frac cp$ are all less than $\frac12.$ Thus $\frac apA + \frac bpB + \frac cpc < \frac12(A+B+C) = \frac12(180^\circ) = 90^\circ.$

For the other inequality, the largest angle of a triangle is always opposite the longest side, and the smallest angle is opposite the shortest side. (That is "geometrically obvious", but I don't offhand know a proof of it.) So the weighted mean $\frac apA + \frac bpB + \frac cpC$ gives the greatest weight to the largest angle and the least weight to the smallest angle, and therefore must be greater than (or equal to) the unweighted mean $\frac13 A + \frac13 B + \frac13 C = \frac13(180^\circ) = 60^\circ.$[/sp]

Nicely done, Opalg. I would like to add a quick comment.

Spoiler

The ratio in the middle of the inequality remains identical for any triangle similar to $\triangle ABC$ as each side $a$, $b$, $c$ will just be replaced by $ka$, $kb$, $kc$ off some factor $k$. So without loss, we can just assume $a+b+c=1$. That makes the problem easier.

If you assume that, $< 90^\circ$ comes from the triangle inequality (to justify each side is less than $\frac 12$), and $60^\circ \leq$ comes from $180 = (A+B+C)(a+b+c)$ and the rearrangement inequality.