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Solutions to Problems; Chapter 7 361

from which the inequality (*) follows, since 0 < w < 1. Equality occurs
C4(u+V)w=v(j(u+v)(l-w)=uew/(l-w)=v/u~a2u=
b2v _ ad = bc.

4.10. The proof follows a modified induction argument on x. The result
clearly holds for x = 0. Suppose 1 < x < 5. Then N(x) = 2, while

x - (x + 1)(x + 5)/12 = (5 - x)(x - 1)/12 > 0

and
(x2 + 6x + 12)/12 - x = [(x - 3)2 + 31112 > 0.
Suppose x 2 6 and that the result has been established for integers up
to x - 6, i.e. that

(x2 - 6x + 5)/12 5 N(x - 6)^5 (x2 - 6x + 12)/12.

Since N(x) = N(x - 6) + 2, the result follows immediately.

4.11. Suppose that A and B are acute. If TA = tan A/2 and v = tan B/2,
then 0 < u, v 5 1 and tanC/2 = cot(A+B)/2 = (l-uv)/(u+v). Denoting
the left side of the required inequality by S, we have that

s = u2+v2+(1-uv)z(u+v)-2

= [(u + v) + (1 - UV)(U + v)--‘12 - 2.

Since 4uv 5 (u + v)~, (uv)(u + v)-’ 2 (u + v)/4. Hence

s 2 [3(u + v)/4 + (u + v)-y2 - 2 > 3 - 2 = 1

by application of the AGM inequality to 3(u + v)/4 and (u + v)-‘.

4.12. The zeros of the quadratic are real iff u2 + 18~ + 9 2 0, i.e. when
u < -9 - 64 or u 2 -9 + Sfi. The sum of the squares of the zeros is
5u2+ 12~ + 5 = 5(u + 6/5)2 - 1115. The overall minimum of this quantity
occurs at u = -615, and the minimum subject to the constraint on u occurs
for the value of u closest to -615, namely u = -9 + Sfi.

4.13.

x+y=m(x/m)+n(y/n)=x/m+...+x/m+y/n+.--+y/n

2 (m + n)[(x/m)m(y/n)“]‘~(m+“)

by the AGM inequality.

Remark. An elegant, albeit contrived, variant on the idea in the solution
is the problem of maximizing x2y subject to x, y 2 0 and

x+y+ d2x2+2xy+3y2 = k.