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360 Answers to Exercises and Solutions to Problems

Since, subject to u + v = 1, uv 5 l/4, it follows that c/a - b/u 2 4(9/2)2 -

1 = 80. Equality occurs everywhere in the above inequalities iff p = q =

r=s=4.

If a = 0, b # 0, then it is required to show that c/b 1 1, i.e. p + q + r 2 1

where p, q, r are the zeros of the now cubic polynomial. But

p+q+r=(p+q+r)(l/p+l/q+l/r)L3.

Finally, if a = b = 0, it must be shown that c 1 0, which is obvious.
4.8. For n = 1, the inequality is (x - 1)2 2 0. Suppose the inequality holds
forn=k-l>l.Then

xk+’ - (k + 1)x + k = x[xk - kx + (k - l)] + k(x - 1)2 > 0

and the result follows by induction.

4.9. Let c = ua3 and d = vb3. Then the given condition can be rewritten

u2a6 + v2b6 = (a2 + b2)3

which reduces to u2w3+ v2(1 - w)~ = 1, where w = u2(u2 + b2)-l. We have
to show that 11-l + v-r 1 1, or equivalently u + v 1 UV.
To get a purchase on the situation, let us examine the special case in
which ad = bc. This is equivalent to vb2 = ua2, so that w = v(u + v)-’ and
the given condition reduces to

u2v3 + v2u3 = (u + v)” e u2v2 = (u + v)” e 1 = (U--l + v-i)?

This suggests that we try to show that

1 = u2w3 + v2(1 - w)3 2 u2[v/(u + v)]3 + V2[(U/(U + v)]”

= u2v2/(u + v)2 = (l/u + 1/v)-?

To this end, observe that

t*>

(u + v)3u2w3 + (u + v)V(l - w)3 - u2v3 - U3V2

= u”[(u + v)3w3 - v3] + v”[(u + v)3(1 - w)3 - U”]

= u2[(u+v)w-v][(u+v)2w2+(u+v)vw+v2]

+ v2[(u + v)(l - w) - U][(U + v)2( 1 - w)2

+ (u + v)u(l - w) + u2]

= [(u + v)w - vJ{u’[(u + v)2w2 + (u + v)vw + v”]

• v2[(u + v)2(1- 2w + w2) + (TJ + v)u(l - w) + U2]}
  = [(u+v)w - v](u + v)[(u2 - v”)(u + v)w2

• v(u + v)(u + 2v)w - vy2u + v)]
  = (u + v)[(u + v)w - v12[(u2 - v2)w + v(2u + v)]
  = (u + v)[(u + v)w - V]“[U”W + 2uv + (1 - w)v2] 1 0,