By Daniel J. Velleman

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**Additional resources for American Mathematical Monthly, volume 117, number 4, April 2010**

**Example text**

2); in other words, r + s ≡ n − 1 (mod 2). Note also that, while F(sin z, cos z) could be rewritten in any number of ways by using sin2 z + cos2 z = 1 (so again the homogeneity of F cannot be essential), the parities of r and s would remain the same. 5) becomes n H (z) = k=1 n = k=1 ∞ F(sin ak , cos ak ) 1 1≤ j ≤n sin(ak − a j ) m=−∞ z − ak − mπ j =k F(sin ak , cos ak ) cot(z − ak ). 1≤ j ≤n sin(ak − a j ) j =k To be able to apply Liouville’s theorem we need G(z) − H (z) to be bounded, so let us also assume for now that r + s < n.

26) Moreover, inequality (23) is then strict except when p = 0 or 1, or the sets {a, b, c} and {x, y, z} coincide. Proof. ) We discuss only hypothesis (25), the necessity of (24) and (26) being established as in Theorem 1. It follows from (23) that the inequality ap + bp + cp 3 1/ p x p + yp + zp 3 ≤ 1/ p (27) is valid when p < 0, and that it reverses direction when 0 < p ≤ 1. 3]. When p → 0− we deduce that (abc)1/3 ≤ (x yz)1/3 , (28) and, when p → 0+ , that (28) is reversed. ) We assume throughout that the sets {a, b, c} and {x, y, z} are disjoint; otherwise they coincide and the assertion of the theorem is trivial.

Suppose no such δ exists. Then for every integer n ≥ 1 we can find a point an ∈ Bd ∗ (x, 1/n) with an ∈ Bd (x, ). Since d ∗ (x, an ) < 1/n we can fix a (1/n)-chain x0 , x1 , . . xk connecting x to an . Thus x0 = x, xk = an , and d(xi , xi+1 ) < 1/n for i < k. Call x0 , x1 , . . , xk−1 , xk , xk−1 , . . , x1 , x0 the (1/n)-circuit. It leads from x to a point outside of Bd (x, ) and then back again with steps of length < 1/n. Now define a sequence yi by starting with the 1-circuit, following it by the (1/2)-circuit, then the (1/3)-circuit, etc.