IMO 2001 SL A6

Origin: KOR | Category: Algebra

Problem

Prove that for all positive real numbers a, b, c, a \sqrt a2 + 8bc + a \sqrt b2 + 8ca + c \sqrt c2 + 8ab \geq1.

Solution

We shall determine a constant k > 0 such that a \sqrt a2 + 8bc \geq ak ak + bk + ck for all a, b, c > 0. (1) This inequality is equivalent to (ak + bk + ck)2 \geqa2k−2(a2 + 8bc), which further reduces to (ak + bk + ck)2 −a2k \geq8a2k−2bc. On the other hand, the AM–GM inequality yields (ak + bk + ck)2 −a2k = (bk + ck)(2ak + bk + ck) \geq8ak/2b3k/4c3k/4, and therefore k = 4/3 is a good choice. Now we have a \sqrt a2 + 8bc + b \sqrt b2 + 8ca + c \sqrt c2 + 8ab \geq a4/3 a4/3 + b4/3 + c4/3 + b4/3 a4/3 + b4/3 + c4/3 + c4/3 a4/3 + b4/3 + c4/3 = 1. Second solution. The numbers x = a \sqrt a2+8bc, y = b \sqrt b2+8ca and z = c \sqrt c2+8ab satisfy f(x, y, z) = 1 x2 −1 1 y2 −1 1 z2 −1 = 83.

Our task is to prove x + y + z \geq1. Since f is decreasing on each of the variables x, y, z, this is the same as proving that x, y, z > 0, x + y + z = 1 implies f(x, y, z) \geq83. However, since x2 −1 = (x+y+z)2−x2 x2 = (2x+y+z)(y+z) x2 , the inequality f(x, y, z) \geq83 becomes (2x + y + z)(x + 2y + z)(x + y + 2z)(y + z)(z + x)(x + y) x2y2z2 \geq83, which follows immediately by the AM–GM inequality. Third solution. We shall prove a more general fact: the inequality a \sqrt a2+kbc + b \sqrt b2+kca + c \sqrt c2+kab \geq \sqrt1+k is true for all a, b, c > 0 if and only if k \geq8. Firstly suppose that k \geq8. Setting x = bc/a2, y = ca/b2, z = ab/c2, we reduce the desired inequality to F(x, y, z) = f(x) + f(y) + f(z) \geq \sqrt 1 + k , where f(t) = \sqrt 1 + kt, (2) for x, y, z > 0 such that xyz = 1. We shall prove (2) using the method of Lagrange multipliers. The boundary of the set D = {(x, y, z) \inR3

| xyz = 1} consists of points (x, y, z) with one of x, y, z being 0 and another one being +\infty. If w.l.o.g. x = 0, then F(x, y, z) \geqf(x) = 1 \geq3/ \sqrt 1 + k. Suppose now that (x, y, z) is a point of local minimum of F on D. There exists \lambda \inR such that (x, y, z) is stationary point of the function F(x, y, z)+\lambdaxyz. Then (x, y, z, \lambda) is a solution to the system f ′(x)+\lambdayz = f ′(y) + \lambdaxz = f ′(z) + \lambdaxy = 0, xyz = 1. Eliminating \lambda gives us xf ′(x) = yf ′(y) = zf ′(z), xyz = 1. (3) The function tf ′(t) = −kt 2(1+kt)3/2 decreases on the interval (0, 2/k] and increases on [2/k, +\infty) because (tf ′(t))′ = k(kt−2) 4(1+kt)5/2 . It follows that two of the numbers x, y, z are equal. If x = y = z, then (1, 1, 1) is the only solution to (3). Suppose that x = y ̸= z. Since (yf ′(y))2 −(zf ′(z))2 = k2(z−y)(k3y2z2−3kyz−y−z) 4(1+ky)3(1+kz)3 , (3) gives us y2z = 1 and k3y2z2−3kyz−y−z =

Eliminating z we obtain an equation in y, k3/y2 −3k/y −y −1/y2 = 0, whose only real solution is y = k −1. Thus (k −1, k −1, 1/(k −1)2) and the cyclic permutations are the only solutions to (3) with x, y, z being not all equal. Since F(k −1, k −1, 1/(k −1)2) = (k + 1)/ \sqrt k2 −k + 1 > F(1, 1, 1) = 1, the inequality (2) follows. For 0 < k < 8 we have that a \sqrt a2+kbc + b \sqrt b2+kca + c \sqrt c2+kab > a \sqrt a2+8bc + b \sqrt b2+8ca + c \sqrt c2+8ab \geq1. If we ﬁx c and let a, b tend to 0, the ﬁrst two sum- mands will tend to 0 while the third will tend to 1. Hence the inequality cannot be improved.