Solutions to Problems; Chapter 7 359
Equality occurs only when 1 - x = x - y = y - z = z = l/4, so that the
only solution is (2, y, z) = (314, l/2, l/4).
4.2. z2 + (1/2)(x2 + y2) - (x + y)z = (1/2)[(z - x)~ + (z - Y)~] 1 0.
4.3.Letthezerosberl,rz,...,rn.Thenrl+r2+...+rn=1andrl’+
r2 -1 +... + r;’ = n2. By the CSB inequality,
n = pPr+/2
i I &i?Tim=n.Thus, equality occurs between the middle members and rz = ri/r,T’ must
be the same for each i.
4.4. Let the zeros be rl, r2,... ,r”. If all the zeros were real, then, by the
AGM inequality,
r: + ri +... + rz 1 n(rfri... rz)““.Expressing the symmetric functions in terms of the coefficients yields a2 -
2b 2 nk2/“, which contradicts the hypothesis. Hence, not all zeros are real.
4.5. x3 + y3 + 23 - x2y - y2z - 222
= x2(x - y) + y2(y - z) - z2(x - z)
= (x2 - z”)(x - y) + (y2 - 22)(y - %)
= (x - y)(x - z)(x + %) + (y - %)2(y + z).If either x 5 y 5 z or x 2 y 2 z, this expression is clearly seen to be
nonnegative and the inequality follows. For the other possible relations
among x, y, Z, a corresponding expression for the difference between the
two sides occurs which makes the desired inequality plain.
4.6. The difference of the two sides is
%[X” + y2 + z2 - xy - yz - ZX] = (z/2)[(x - y)2 + (y - %)2 + (z - x)2].4.7. First, consider the case that a # 0. Let p, q, r, s be the (positive)
zeros of at4 -bt3+ct2-t+l,andletu=l/p+l/qandv=l/r+l/s.
Then u + v = 1, p + q = pqu and r + s = rsv. It is required to show that
c/a - b/u 2 80.
Nowc/u-b/u=(p+q)(r+s)+pq+rs-(p+q)-(r+s)=pqrsuv+
pq(1 - u) + rs(l - v) = pqrsuv + pqv + rsu. By the AGM inequality,
(m-l 5 (p-l + q-‘)/2, whence pq 2 4,~~~. Similarly, rs > 4vm2. Hence
c/u - b/a 2 16(uv)-’ + ~(uv)-~(u” + v3)
= 16(uv)-1 + ~(uv)-~(u~ - uv + v2) (since u + v = 1)
= 4[3(uv)-1 + (v-2 + u-“)I
= 4[(uv)-’ + (u-l + v-l)“] = 4[(uv)-’ + (uv)-2]
= 4(1/2 + 11~~)~ - 1.