University of Central Florida
MAC 2220
Solution to H3 Math 2015 Lecture Test 1
1. (i) Note that if a > 0, then the quadratic expression ax2 + 2bx + c is
nonnegative if and only if the discriminant is nonpositive, i.e. (2b)2 -
4(a)(c) ≤ 0, that is if and only if b2 ≤ ac.
Otherwise, note that
ax2 + 2bx + c = a[(x + b
a
)2 + c
a
-
b2
a2 ] = a(x + a b )2 + c - b a
...[Show More]
Solution to H3 Math 2015 Lecture Test 1
1. (i) Note that if a > 0, then the quadratic expression ax2 + 2bx + c is
nonnegative if and only if the discriminant is nonpositive, i.e. (2b)2 -
4(a)(c) ≤ 0, that is if and only if b2 ≤ ac.
Otherwise, note that
ax2 + 2bx + c = a[(x + b
a
)2 + c
a
-
b2
a2 ] = a(x + a b )2 + c - b a 2 :
This expression is nonnegative if and only if c - b a 2 ≥ 0, i.e. b2 ≤ ac
since a(x + a b )2 ≥ 0 as a > 0.
(ii) Note that
nX i
=1
(aix + bi)2 =
nX i
=1
a2
i ! x2 + 2 X i=1 n aibi! x + X i=1 n b2 i ! :
Since Pn i=1(aix + bi)2 ≥ 0, using the result in (i), we have
nX i
=1
aibi!2 ≤ X i=1 n a2 i ! X i=1 n b2 i ! :
This is also known as the Cauchy-Schwarz inequality.
(iii) Note that with a1 = u; a2 = v; a3 = w and b1 = v; b2 = w; b3 = u
we have by (ii)
(uv + vw + wu)2 ≤ (u2 + v2 + w2)2
which implies
uv + vw + wu ≤ u2 + v2 + w2; (1)
since u; v; w are positive reals. In addition,
(u + v + w)2 = u2 + v2 + w2 + 2(uv + vw + wu) ≥ 3(uv + vw + wu) ≥ 36
using the above inequality (1) and the condition given. Thus
u + v + w ≥ 6
since u; v; w are positive reals.
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