Baltic Way 2011 · Shortlist problem

Geometry

Let $a$ , $b$ , and $c$ be the lengths of the sides of a triangle, $R$ the radius of its circumcircle, and $r$ the radius of its incircle. Prove that

\frac{Rr}{(a+b+c)^2} \le \frac{1}{54}.

When you’re ready

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Review

Geometric inequalities · Triangles and centers

Solution

We use the identities $2S = (a+b+c)r$ and $\frac{abc}{4S} = R$ , where $S$ denotes the area of the triangle. Multiplying them we obtain

\frac{abc}{2} = Rr(a+b+c) = \frac{Rr}{(a+b+c)^2} \cdot (a+b+c)^3.

It remains to show that $(a+b+c)^3 \ge 27abc$ . This follows directly from AM-GM inequality.