Question 2

Let $$a$$, $$b$$, $$c$$ be the lengths of the sides of a triangle $$ABC$$. Let $$\alpha$$, $$\beta$$, $$\gamma$$ be positive integers and $$p = \beta a + \gamma b + \alpha c$$, $$q = \gamma a + \alpha b + \beta c$$ and $$r = \alpha a + \beta b + \gamma c$$. Then $$p$$, $$q$$, $$r$$ are lengths of the sides of a triangle

Solution

Grouping by $$\alpha$$, $$\beta$$, $$\gamma$$, $$p + q - r = \alpha(b + c - a) + \beta(c + a - b) + \gamma(a + b - c)$$. Each bracket is positive by the triangle inequality and $$\alpha$$, $$\beta$$, $$\gamma$$ are positive, so $$p + q > r$$. The other two inequalities follow in the same way, so $$p$$, $$q$$, $$r$$ always form a triangle.

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