(^292) Answers to Exercises and Solutions to Problems
8.12. Observe that
0 = -w7-x7-y7-%7=(x+y+%)7-x7-y7-%7
= 7(x + y)(y + %)(% + x)[(x” + y2 + %2 + xy + y% + %x)2
- ZY%(Z + y + %)I.
Now,
4(x2 + Y2+~2+xY+Y~+%2)2+4xY%(Z+Y+%)
= [(x + y)2 + (y + %)” + (% + x)“]” - 4xyzw
= [(w+“)2+(w+x)2+(w+y)2]2-4xy%w
= [3w2 + 2w(x + y + %) + x2 + y2 + %“I” - 4XY%W
= [w2 + x2 + y2 + t212 - 4xyzw
1 [2]wx] + 2]yz]]2 - 4xyzw
1 [4( lwxyz ])‘/2]2 - 4xyzw
1 12]xyzw] 10
by a double application of the Arithmetic-Geometric Mean Inequality (Ex-
ercise 1.2.17), with equality iff Ix] = ]y] = ]%I = ]w] = 0. Otherwise, at least
one of x + y, y + %, % + x vanishes. In any case, w(w + Z)(W + y)(w + %)
assumes only the value 0.
8.13. This follows directly from Exercise 2.12. A generalization to four
variables appears in Crux Mathematicorum 2 (1976), 180.
8.14. (a) Using the third equation to eliminate c from the other two yields
=(x2%2 - xy?z) + b(x# + y2z2) = 0
a(x2.z2 + x3y) + b(y2z2 - x2yz) = 0.
Eliminating a and taking account of bxy # 0 leads to
x3%3 + y323 + x3y3 + x2y2t.2 = 0.
(b) Determining abc by multiplying the right sides gives
abc = 2abc + a2bx2zB2 + a2cx2yB2 + b2cxe2y2 + ab2y2zs2 - a~~,z’y-~ + bc2x-2r2.
Multiplying by zy% gives
0 = abcxyz + a2z3(byz-’ + WY-‘) + b2y3(crx-1 + axz-‘)
f c2z3(axy-’ + byx-I)
= abcxyz + a3x3 + b3$ + c3z3.