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Solutions to the Exercises
Solution for Exercise 12.7. Pitman solves his Problem 3.1.24 by first expanding 1 = (x + y + z)3 and then noting that it suffices to show Q = x2 y + x2 z + y 2 x + y 2 z + z 2 x + z 2 y ≤ 1/4
when x + y + z = 1.
If we write Q = x{x(y + z)} + y{y(x + z)} + z{z(x + y)}, then it now suffices to notice that each of the three braced expressions is bounded below by 1/4 by the AM-GM inequality. Other solutions can be based on the homogenization trick of Exercise 12.3, or Schur’s inequality (page 83), or the reduction devices of Exercise 12.6. Solution for Exercise 12.8. This elementary (but very useful!) inequality serves as a reminder that symmetry is often the key to successful telescoping. Here the telescoping identity a1 a2 · · · an − b1 b2 · · · bn =
n
a1 · · · aj−1 (aj − bj )bj+1 · · · bn
j=1
makes the Weierstrass inequality immediate. Naturally, generalizations of this identity lead one to more elaborate versions of Weierstrass inequality. Chapter 13: Majorization and Schur Convexity Solution for Exercise 13.1. From each of the representations a 1/2 1/3 1/6 x a 0 1/2 1/2 x b = 1/3 2/3 b = 1/2 1/6 1/3 y 0 y c 1/6 0 5/6 z c 1/2 1/3 1/6 z one gets (a, b, c) ≺ (x, y, z). The inequalities of the exercise then follow from the Schur concavity of the map (x, y, z) → xyz and the Schur convexity of the map (x, y, z) → 1/x5 + 1/y 5 + 1/z 5 . Solution for Exercise 13.2. If we set s = (x1 + x2 + · · · + xn )/k we have (x1 , x2 , . . . , xn ) ≺ (s, s, . . . , s, 0, 0, . . . , 0) when we take k copies of s. Thus, for convex φ : [0, ∞) → R, Schur’s majorization inequality (13.18) gives us φ(x1 ) + φ(x2 ) + · · · + φ(xn ) ≤ (n − k)φ(0) + kφ(s), and we can set φ(x) = 1/(1 + x) to obtain the bound (13.21). Solution for Exercise 13.3. If one sets µ + δ/m for 1 ≤ k ≤ m yk = µ − δ/(n − m) for m < k ≤ n where µ = (x1 +x2 +· · ·+xn )/n, then from the condition (13.22) it follows