UI Putnam Training Sessions
Problem Set 10: Inequalities II
http://www.math.illinois.edu/contests.html
The classical inequalities: AGM and Cauchy
• The Arithmetic-Geometric Mean (AGM) Inequality
(a1 a2 . . . an )1/n ≤
1
(a1 + a2 + · · · + an )
n
Here the ai are arbitrary nonnegative real numbers. Equality holds if and only if the ai ’s are equal.
• Cauchy’s Inequality
n
X
!2
ai bi
n
X
≤
i=1
!
a2i
i=1
n
X
!
b2i
i=1
Equality holds if and only if the ai ’s are bi ’s are proportional to each other, i.e., if ai = λbi for all i
or bi = λai for all i, for some λ.
Advanced topic: Jensen’s inequality
Jensen’s inequality states that if f (x) is a convex (i.e., concave up) function, then for any real numbers
x1 , x2 , t with 0 ≤ t ≤ 1,
f (tx1 + (1 − t)x2 ) ≤ tf (x1 ) + (1 − t)f (x2 ) .
Geometrically, this simply says that any point on the graph of f between x1 and x2 lies below the
corresponding point on the line segment connecting f (x1 ) and f (x2 ). (Draw a picture!)
More generally, if f is convex, then given any real numbers x1 , . . . , xn and “weights” p1 , . . . , pn
satisfying 0 ≤ pi ≤ 1 and p1 + · · · + pn = 1, we have
f
n
X
!
p i xi
≤
i=1
n
X
pi f (xi )
i=1
If f is concave (such as f (x) = log x), then the reverse inequality holds:
n
X
pi f (xi ) ≤ f
i=1
n
X
!
p i xi
i=1
Remark: The standard way to check convexity is via the second derivative: If f 00 (x) is positive, then f
is convex; if it is negative, then f is concave.
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UI Putnam Training
Problem Set 10: Inequalities II
Fall 2016
The Problems
1. More cool applications of AGM. Try to find a one or two line solution using AGM.
(a) A calculus problem without calculus. The following is a standard Calculus III problem;
try to find a slick solution using AGM, but no calculus!
Determine the maximal volume of a rectangular package whose “girth” is at most 6,
and find the dimensions of an optimal box with girth 6 that achieves this volume.
(The girth of a box with sides x, y, z is defined as x + 2y + 2z.)
(b) The Harmonic-Geometric Mean Inequality: Show that, for any positive real numbers
a1 , . . . , an ,
n
1/n
.
1
1 ≤ (a1 . . . an )
+
·
·
·
+
a1
an
(The left side is the harmonic mean of the numbers a1 , . . . , an . The inequality shows that
the harmonic mean is always less than or equal to the geometric mean.)
(c) (UI Undergraduate Math Contest 1996) Let a, b, c be real numbers > 1, and let
S = loga (bc) + logb (ca) + logc (ab),
where logb x denotes the base b logarithm of x. Find, with proof, the smallest possible value
of S.
(d) (Virginia Tech Math Contest 2002) Let {an }n≥1 be an infinite sequence with an > 0 P
for all n.
1/n
Let bn denote the geometric
mean of a1 , . . . , an , that is, bn = (a1 . . . an ) . Suppose ∞
n=1 an
P∞
is convergent. Prove that n=1 b2n is also convergent.
Remark: P
A much harder version of this problem is to show that, under the same assumptions,
the series ∞
n=1 bn converges..
2
UI Putnam Training
Problem Set 10: Inequalities II
Fall 2016
2. Applications of Jensen’s Inequality.
(a) Proof of the AGM Inequality: Prove the AGM inequality by applying Jensen’s Inequality
to an appropriate concave function.
(b) AGM Inequality with Weights: Prove the following generalized version of the AGM inequality:
ap11 a2p2 . . . apnn ≤ p1 a1 + p2 a2 + · · · + pn an ,
where p1 , . . . , pn are nonnegative real numbers satisfying p1 + · · · + pn = 1, and a1 , . . . , an are
arbitrary positive real numbers.
(The standard AGM inequality corresponds to the case when the pi ’s are all equal to 1/n.)
(c) Let x, y, z be real numbers satisfying 0 < x, y, z < 1 and x + y + z = 1. Show that
xx y y z z ≤ x2 + y 2 + z 2 .
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UI Putnam Training
Problem Set 10: Inequalities II
Fall 2016
3. Miscellaneous inequality problems. Here are some harder problems. Some of these can be
done with clever applications of standard inequalities, others require “ad hoc” methods.
(a) (A2, Putnam 2003; U of I Mock Putnam Exam 2011) Let a1 , . . . , an , b1 , . . . , bn be positive real
numbers. Prove that
(a1 . . . an )1/n + (b1 . . . bn )1/n ≤ (a1 + b1 )1/n (a2 + b2 )1/n · · · (an + bn )1/n .
(b) Let a1 , a2 , . . . , an be a permutation of the numbers 1, 2, . . . , n. Show that the sum
S = 1 · a1 + 2 · a2 + · · · + n · an
attains its maximal value for the identity permutation, i.e., when ak = k for all k.
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UI Putnam Training
Problem Set 10: Inequalities II
Fall 2016
Challenge Problem of the Week
This week’s problem is really hard; in fact, it may be the most challenging problem we have had so
far. It is a (much) harder version of the Virginia Tech Contest problem stated earlier. The problem is
deceptively easy to state, and it looks very much “doable” with a standard application of AGM, but if
one tries that, one runs into seemingly unsurmountable difficulties ...
Convergence of the geometric mean series. Let an be positive real numbers,
and letP
bn = (a1 a2 . . . an )1/n be the geometric mean
P∞of the first n terms ai . Show
a
converges,
then
so
does
the
series
that if ∞
n=1 bn .
n=1 n
Happy Problemsolving!
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