IMO 1973 SL 11

Origin: SWE

Problem

Determine the minimum of a2 + b2 if a and b are real numbers for which the equation x4 + ax3 + bx2 + ax + 1 = 0 has at least one real solution.

Solution

Putting x + 1 x = t we also get x2 + x2 = t2 −2, and the given equation reduces to t2 + at + b −2 = 0. Since x = t\pm \sqrt t2−4 , x will be real if and only if |t| \geq2, t \inR. Thus we need the minimum value of a2 + b2 under the condition at + b = −(t2 −2), |t| \geq2. However, by the Cauchy–Schwarz inequality we have

(a2 + b2)(t2 + 1) \geq(at + b)2 = (t2 −2)2. It follows that a2 + b2 \geqh(t) = (t2−2)2 t2+1 . Since h(t) = (t2 + 1) + t2+1 −6 is increasing for t \geq2, we conclude that a2 + b2 \geqh(2) = 4 5. The cases of equality are easy to examine: These are a = \pm 4 5 and b = −2 5. Second solution. In fact, there was no need for considering x = t+1/t. By the Cauchy–Schwarz inequality we have (a2 + 2b2 + a2)(x6 + x4/2 + x2) \geq (ax3 + bx2 + ax)2 = (x4 + 1)2. Hence a2 + b2 \geq (x4 + 1)2 2x6 + x4 + 2x2 \geq4 5, with equality for x = 1.