IMO 1977 LL ROM38

Origin: ROM

Problem

Let mj > 0 for j = 1, 2, . . ., n and a1 \leq\cdot \cdot \cdot \leqan < b1 \leq\cdot \cdot \cdot \leq bn < c1 \leq\cdot \cdot \cdot \leqcn be real numbers. Prove: ⎡ ⎣ n  j=1 mj(aj + bj + cj) ⎤ ⎦

3 ⎛ ⎝ n  j=1 mj ⎞ ⎠ ⎡ ⎣ n  j=1 mj(ajbj + bjcj + cjaj) ⎤ ⎦.

Solution

The condition says that the quadratic equation f(x) = 0 has distinct real solutions, where f(x) = 3x2 n  j=1 mj −2x n  j=1 mj(aj + bj + cj) + n  j=1 mj(ajbj + bjcj + cjaj). It is easy to verify that the function f is the derivative of F(x) = n  j=1 mj(x −aj)(x −bj)(x −cj). Since F(a1) \leq0 \leqF(an), F(b1) \leq0 \leqF(bn) and F(c1) \leq0 \leqF(cn), F(x) has three distinct real roots, and hence by Rolle’s theorem its deriva- tive f(x) has two distinct real roots.