### 43rd Putnam 1982

Problem B6

Let A(a, b, c) be the area of a triangle with sides a, b, c. Let f(a, b, c) = √A(a, b, c). Prove that for any two triangles with sides a, b, c and a', b', c' we have f(a, b, c) + f(a',b',c') ≤ f(a + a', b + b', c + c'). When do we have equality?

Solution

Answer: equality iff the triangles are similar (a/a' = b/b' = c/c').

We prove first that for any positive reals A, B, A', B' we have √(AB) + √(A'B') ≤ √( (A + A')(B + B') ) with equality iff A/A' = B/B' . We have (√(AB') - √(A'B) )2 ≥ 0, with equality iff A/A' = B/B' . Hence 2√(ABA'B') ≤ AB' + A'B. Hence (AB + A'B') + 2√(ABA'B') ≤ (AB + A'B') + AB' + A'B. Now the lhs is (√(AB) + √(A'B') )2. The rhs is (A + A')(B + B'). So taking the square root, we have √(AB) + √(A'B') ≤ √( (A + A')(B + B') ) with equality iff A/A' = B/B' .

Now set A = √(ab), B = √(cd), A' = √(a'b'), B' = √(c'd'). Then the relation becomes (abcd)1/4 + (a'b'c'd')1/4 ≤ √( (√(ab) + √(a'b') )(√(cd) + √(c'd') ). Applying the relation again to each of the inner parentheses on the rhs, we get (√(ab) + √(a'b') )(√(cd) + √(c'd') ≤ √( (a + a')(b + b')(c + c')(d + d') ). So we arrive finally at (abcd)1/4 + (a'b'c'd')1/4 ≤ ( (a + a')(b + b')(c + c')(d + d') )1/4. We have equality iff A/A' = B/B', a/a' = b/b' and c/c' = d/d' and hence iff a/a' = b/b' = c/c' = d/d'.

Applying this result to Heron's formula gives the required result immediately, with equality iff a/a' = b/b' = c/c'.