Let $ABC$ be a triangle. Show that $$ 60^o \le \frac{aA+bB+cC}{a+b+c} \le 90^o$$ where $A$ is the measure in degrees of the angle corresponding to the vertex $A$ and $a$ the side opposite to it ($B$, $b$, $C$ and $c$ are defined similarly).
I saw this problem in a Facebook group and couldn't figure out the solution. It was a long time ago, so there is no way I can find that post again (only took a screenshot of the problem).
I have no idea where to start. My thoughts about the problem:
- $a+b+c$ is the perimeter of the triangle.
- The quantity $\frac{aA+bB+cC}{a+b+c}$ is the weighted average angle with sides as weights.
- The formula is obvious for an equilateral triangle.
- Since $\frac{aA+bB+cC}{a+b+c}=60^o$ is obtained for an equilateral triangle, I was thinking using Lagrange multipliers method with the Law of Cosines, Law of Sines and $A+B+C=180^0$ has constraints but the computations seem complicated.
Any idea?
Let's start with the right inequality. Notice that $a,b,c$ satisfy $a+b+c>2a, a+b+c>2b, a+b+c>2c$ due to the triangle inequality. Multiply these inequalities respectively with $\alpha, \beta,\gamma$ $$(\alpha+\beta+\gamma)\cdot (a+b+c)>2\cdot(a\alpha+b\beta+c\gamma)\iff90^\circ=\frac{\alpha+\beta+\gamma}2>\frac{a\alpha+b\beta+c\gamma}{a+b+c}$$
For the second one, suppose wlog $a\geqslant b\geqslant c$. This implies, as it is well-known, that $ \alpha\geqslant \beta\geqslant \gamma$ so you can use Chebyshev's inequality as follows $$\frac{a\alpha+b\beta+c\gamma}3\geqslant \frac{(a+b+c)}3\cdot\frac{(\alpha+\beta+\gamma)}3\iff \frac{a\alpha+b\beta+c\gamma}{a+b+c}\geqslant \frac{\alpha+\beta+\gamma}3=60^\circ$$