IMO 1977 LL VIE58

Origin: VIE

Problem

Prove that for every triangle the following inequality holds: ab + bc + ca 4S \geqcot \pi 6 , where a, b, c are lengths of the sides and S is the area of the triangle.

Solution

The following inequality (Finsler and Hadwiger, 1938) is sharper than the one we have to prove: 2ab + 2bc + 2ca −a2 −b2 −c2 \geq4S \sqrt 3. (1) First proof. Let us set 2x = b + c −a, 2y = c + a −b, 2z = a + b −c.

Then x, y, z > 0 and the inequality (1) becomes y2z2 + z2x2 + x2y2 \geqxyz(x + y + z), which is equivalent to the obvious inequality (xy −yz)2 + (yz −zx)2 + (zx −xy)2 \geq0. Second proof. Using the known relations for a triangle a2 + b2 + c2 = 2s2 −2r2 −8rR, ab + bc + ca = s2 + r2 + 4rR, S = rs, where r and R are the radii of the incircle and the circumcircle, s the semiperimeter and S the area, we can transform (1) into s \sqrt 3 \leq4R + r. The last inequality is a consequence of the inequalities 2r \leqR and s2 \leq 4R2 + 4Rr + 3r2, where the last one follows from the equality HI2 = 4R2 + 4Rr + 3r2 −s2 (H and I being the orthocenter and the incenter of the triangle).