Find the largest and smallest values of w(w + x)(w + y)(w + z) for reals w, x, y, z such that w + x + y + z = 0 and w7 + x7 + y7 + z7 = 0.
Solution
Solution by Demetres Christofides
Answer: both 0.
We have that (w+x)7 - 7wx(w+x)(w2+wx+x2)2 = w7 + x7. Similarly for y and z. So adding gives wx(w+x)(w2+wx+x2)2 + yz(y+z)(y2+yz+z2)2 = 0. Dividing by w+x = -(y+z), we get wx(w2+wx+x2)2 = yz(y2+yz+z2)2.
We have w2 + wx + x2 = (w+x)2 - wx, and y2 + yz + z2 = (y+z)2 - yz = (w+x)2 - yz. Hence (wx - yz)(w+x)4 - 2(wx)2(w+x)2 + 2(yz)2(w+x)2 + (wx)3 - (yz)3 = 0. So either (wx - yz) = 0 or (w+x)4 - 2(wx + yz)(w+x)2 + ( (wx)2 + wxyz + (yz)2) = 0.
If wx = yz, then since also w + x = -(y + z), we have that w and x are the roots of the quadratic t2 + (y+z) t + yz. Solving, w, x = -y, -z in some order. So (w + y)(w + z) = 0.
So suppose wx - yz is non-zero. Then (w+x)4 - 2(wx + yz)(w+x)2 + ( (wx)2 + wxyz + (yz)2) = 0, or ( (w+x)2 - (wx + yz) )2 = wxyz (*). Since (w+x)2 = (y+z)2, we have 2( (w+x)2 - (wx + yz) ) = ( (w+x)2 - 2wx + (y+z)2 - 2yz) = (w2 + x2 + y2 + z2). So (*) becomes (w2 + x2 + y2 + z2)2 = 4wxyz. But by AM/GM we have (w2 + x2 + y2 + z2) > 4(w2x2y2z2)1/4 and hence (w2 + x2 + y2 + z2)2 >= 16wxyz. So 4wxyz >= 16wxyz. Also 4wxyz is non-negative, since it is a square. So we must have wxyz = 0. Hence also (w2 + x2 + y2 + z2) = 0, so w = x = y = z = 0 and we have a special case of wx = yz after all.
So the only possible value of w(w + x)(w + y)(w + z) is zero.
© John Scholes
jscholes@kalva.demon.co.uk
11 Sep 2002