I came across the following exercise
If a side of a triangle is less than the Averages of the two others sides, then the opposite angle is less than the average of the two other angles.
Can anyone help me to prove this statement
I failed to apply Alkashi theorem here.
Simple by drawing the involved loci:
The first constraint ensures that $A$ lies outside the depicted ellipse, hence it also lies outside the depicted circles. In algebraic terms, $a<\frac{b+c}{2}$ implies $$\cos(A)=\frac{b^2+c^2-a^2}{2bc}>\frac{1}{2}$$ hence $A>60^\circ$ and $A>\frac{B+C}{2}=90^\circ-\frac{A}{2}$.