2. KöMaL problem F. 3097.

A convex quadrilateral ABCD

is inscribed in a unit circle. Its sides satisfy the inequality
AB · BC · CD · DA ≥ 4. Prove that ABCD is a square.

B A

2α
“Quickie” Inequalities 2β 2γ2δ

A Murray S. Klamkin†A D
C
“Quickie” problems first appeared in the March 1980 is- Let the angles subtended by the four sides from the center
sue of Mathematics Magazine. They were originated by the be 2α, 2β, 2γ, and 2δ (see figure above). Then AB = 2 sin α,
late Charles W. Trigg, a prolific problem proposer and solver BC = 2 sin β, CD = 2 sin γ and CD = 2 sin δ where α + β +
who was then the Problem Editor. Many of the first good γ + δ = π, π > α, β, γ, δ > 0. Since ln(sin x) is concave,
Quickie proposals were due to the late Leo Moser (who in-
cidentally was a member of the University of Alberta Math-
 π
ln(sin α) + ln(sin β) + ln(sin γ) + ln(sin δ) ≤ 4 ln sin
ematics Department and subsequently its chairman). These 4
Quickie problems are even now still a popular part of the jour-
nal. Also Quickies have proliferated to the problem sections or AB · BC · CD · DA ≤ 4. Hence the product is exactly 4
of Crux Mathematicorum, Math Horizons, SIAM Review and and α = β = γ = δ = π4 so ABCD is a square.
Mathematical Intelligencer (unfortunately, no longer in the
latter two journals). Exercise. Of all convex n-gons inscribed in a unit circle,
determine the maximum of the product of its n sides.
Trigg noted that some problems will be solved by laborious
methods but with proper insight1 may be disposed of with 3. KöMaL problem F. 3238. Prove that the inequality
dispatch. Hence the name “Quickie”. √
p p p 3 2
The probability that two random numbers are equal is a2 + (1 − b)2 + b2 + (1 − c)2 + c2 + (1 − a)2 ≥
zero. It follows that there are more inequalities than equa- 2
tions. Consequently, the study of inequalities are important holds for arbitrary real numbers a, b, c.
throughout mathematics. In past issues of π in the Sky, De- By Minkowski’s Inequality, the sum of the three radicals is
cember 2001, September 2002, Professor Hrimiuc has pro- p
vided some good notes on inequalities and we shall be refer- grater or equal than (a + b + c)2 + (3 − a − b − c)2 . Then
ring to some of them. by the power mean inequality or else letting a + b + c = x,
the expression under√ the radical is 2(x − 3/2)2 + 9/2, so the
Here we illustrate 16 Quickie inequalities and after each minimum value is 3 3 .
2
one we include for the interested reader an exercise that can
be solved in a related manner. Exercise. Determine the minimum value of
Our first example will set the stage for our Quickie Inequal- {x3 + (c − y)3 + a3 }1/3 + {y 3 + b3 + (d − x)3 }1/3 ,
ities.
where a, b, c, d are given positive numbers and x, y ≥ 0.
1. There have been very many derivations published giving
the formulas for the distance from a point to a line and a 4. Determine the maximum and minimum z coordinates of
plane. Here is a Quickie derivation for the distance from the the surface
point (h, k, l) to the plane ax + by + cz + d = 0 in E3 . Here
we want to find the minimum value of [(x − h)2 + (y − k)2 + 5x2 + 10y 2 + 2z 2 + 10xy − 2yz + 2zx − 8z = 0 in E3 .
(z − l)2 ]1/2 where (x, y, z) is a point of the given plane. By One method would be to use Lagrange Multipliers. Another
Cauchy’s Inequality, more elementary method would be to use discriminants of
quadratic equations since if z = h is the maximum, the inter-
[(x − h)2 + (y − k)2 + (z − `)2 ]1/2 [a2 + b2 + c2 ]1/2 section of the plane z = h with the quadric must be a single
≥ |a(x − h) + b(y − k) + c(z − `)| point. Even simpler is to express the quadric that is an ellip-
soid as a sum of squares, i.e., (2x+y)2 +(x−y+z)2 +(z−4)2 =
or 16. Hence max z = 8 and min z = 0.
min[(x − h)2 + (y − k)2 + (z − `)2 ]1/2 Exercise. Determine the maximum value of y 2 and z 2 where
2 2
= |ah + bk + c + d|/[a + b + c ] 2 1/2
. x, y, z are real and satisfy

Exercise. Determine the distance from the point (h, k) to (y − z)2 + (z − x)2 + (x − y)2 + x2 = a2 .
the line ax + by + c = 0.
5. Let ar = (br + br+1 + br+2 )/br+1 where b1 , b2 , . . . , bn > 0
†Murray S. Klamkin is a professor emeritus at the University of and br+n = br . Determine the minimum value of
Alberta √ √ √
1 and appropriate knowledge-MSK 3
a1 + 3 a2 + · · · + 3 an .

26
 Even more generally, let 7. Prove the identity
xj = x1j + x2j + · · · + xmj , j = 1, 2, · · · , n, u(v − w)5 + u5 (v − w) + v(w − u)5 + v 5 (w − u)
n
Y +w(u − v)5 + w5 (u − v) = −10uvw(u − v)(v − w)(w − u),
where all xij > 0 and xij = Pin , i = 1, 2, · · · , m. Then
j=1 and from this obtain the triangle inequality
√ √ √ p
S≡ r
x1 + r x2 + · · · + r xn ≥ n r P1 + P2 + · · · + Pm . aR1 (a4 + R14 ) + bR2 (b4 + R24 ) + cR3 (c4 + R34 ) ≥ 10abcR1 R2 R3

We first use the Arithmetic–Geometric Mean Inequality to (with the same notation as in Problem 6).
get
S ≥ n(x1 x2 · · · xn )1/rn . The identity is a 6th degree polynomial. The left hand side
vanishes for u = 0, v = 0, w = 0, u = v, v = w, and w = u.
Then applying Holder’s Inequality we are done. There is Hence the right hand side equals kuvw(u − v)(v − w)(w − u),
equality if and only if xij = xjk for all i, j, k. where k is a constant. On comparing the coefficients of uv 3 w2
on both sides, k = −1.
The given inequality corresponds to the special case where
r√= m = 3, P1 = P2 = P3 = 1, so that the minimum value is Now, let u, v, w denote complex numbers representing the
n 3 3. vectors from the point P to the respective vertices A, B, C.
Taking the absolute values of the both sides of the identity
The inequalities here are extensions of problem #M1277, and using the triangle inequality |z1 + z2 | ≤ |z1 | + |z2 |, we
Kvant, 1991, which was to show that obtain the desired triangle inequality.
n
X √ Exercise. Referring to Problem 6, prove that
{ai + ai+1 )/ai+2 }1/2 ≥ n 2.
i=1 aR1 R10 + bR2 R20 + cR3 R30 ≥ abc, where R10 , R20 , R30

are the distances from another point Q to the respective ver-

Exercise. Gy. 2887, KöMal. The positive numbers a1 , a2 , tices A, B, C.
. . . , an add up to 1. Prove the following inequality:
8. Determine the maximum and minimum values of
(1 + 1/a1 )(1 + 1/a2 ) · · · (1 + 1/an ) ≥ (n + 1)n . x2 + y 2 + z 2 subject to the constraint x2 +y 2 +z 2 +2xyz = 1.

Since it is known that cos2 α + cos2 β + cos2 γ +

2 cos α cos β cos γ = 1 is a triangle identity, we let x = cos α,
6. If a, b, c are sides of a triangle ABC and R1 , R2 , R3 are the y = cos β, and z = cos γ where α + β + γ = π and π ≥ α,
distances from a point P in plane of ABC to the respective β, γ ≥ 0. Clearly the maximum of cos2 α + cos2 β + cos2 γ
vertices A, B, C. Prove that is 3 and is taken on for (x, y, z) = (1, 1, −1) and permutations
thereof. For the minimum (using the above),
aR12 + bR22 + cR32 ≥ abc.
3
This is a polar moment of inertia inequality and is a special cos2 α + cos2 β + cos2 γ = 3 − (sin2 α + sin2 β + sin2 γ) ≥ .
case of the more general inequality 4

~ + yB
(xA ~ + z C)
~ 2 ≥ 0,
Exercise. Determine the maximum of
~ B,
where A, ~ C~ are vectors from P to the respective vertices ( n ) n q 
A, B, C. Expanding out the square, we get X X 
xi a2i − x2i ,
~ ·C
x2 R12 + y 2 R22 + z 2 R32 + 2yz B ~ + 2zxC
~ ·A
~ + 2xy A
~ · B.
~ i=1

j=1


Since 2B~ ·C~ = R2 + R2 − a2 , etc., the general polar moment where ai ≥ xi ≥ 0.

2 3
of inertia inequality reduces to
9. Problem # 2, Final Round 21st Austrian Mathematical
(x + y + z)(xR12 + yR22 + zR32 ) 2 2
≥ yza + zxb + xyc . 2 Olympiad. Show that for all natural numbers n > 2,
q √ √
Many triangle inequalities are special cases since x, y, z are 3 4 √
2 3 4 · · · n n < 2.
arbitrary real numbers. In particular by letting (x, y, z) =
(a, b, c), we get our starting inequality. Letting P be the cir-
cumcenter and x = y = z, we get 9R2 ≥ a2 + b2 + c2 or
equivalently sin2 A + sin2 B + sin2 C ≤ 94 . Here we get a better upper bound. If P denotes the left
hand side, then
Exercise. Prove that
ln 2 ln 3 ln n
aR2 R3 + bR3 R1 + cR1 R2 ≥ abc. ln P = + + ··· + .
2! 3! n!

27
 ln x and similar formulas for the other medians. It now follows
Since is a decreasing function for x ≥ e,
x that 9m2a − 9m2b = 4(a2 + f 2 ) − 4(b2 + e2 ). It is now possible
  to have ma = mb with their respective face areas unequal, so
ln 2 ln 3 1 1 1 that the longest median is not one to the smallest face area.
ln P < + + + + ...
2! 3 2! 3! 4! The valid analogy is that the longest median is the one to
ln 2 ln 3 the face for which the sum of the squares of its edges is the
= + (e − 2) ≈ 1.7592. smallest, and the shortest median is the one to the face for
2! 3 which the sum of the squares of its edges is the largest.

Exercise. Determine a good lower bound for P . Exercise. Prove that the four medians of a tetrahedron are
possible sides of a quadrilateral.
10. Prove that for any distinct real numbers a, b,
13. a, b, c, d are positive numbers such that a5 +b5 +c5 +d5 =
b
e −e a
b+a e5 . Can an + bn + cn + dn = en for any number n > 5?
>e 2 .
b−a Let St = xt1 + xt2 + · · · + xtn where the xi ≥ 0. A known
result [2] is that the sum St of order t, defined by St = (St )1/t
This is a special case of the following result due to decreases steadily from min xi to 0 as t increases from −∞
J. Hadamard [1]: If a function f is differentiable, and its to 0− , and decreases steadily from ∞ to max xi as t increases
derivative is an increasing function on a closed interval [r, s], from 0+ to +∞. Consequently, there is no such n.
then for all x1 , x2 ∈ (r, s) (x1 6= x2 ), then
Z x2   n
f (x)dx x2 + x1
X
>f . Exercise. Prove that ST ≤ α1 Sti for arbitrary ti > 0 and
x1 x2 − x1 2 i=1
Xn Xn
for α1 > 0, α1 = 1 and T = α1 ti .
Letting f (x) = ex , we get the desired result. i=1 i=1

Exercise. Prove that 14. Prove that

2 2 (b+a)2 xt+1 y t+1 z t+1
eb − ea > (b2 − a2 )e 4 . + + ≥x+y+z
yt zt xt
where x, y, z > 0 and t ≥ 0.
11. Prove that
Let
cosh(y−z)+cosh(z−x)+cosh(x−y) ≥ cosh x+cosh y+cosh z    1
 t+1
t+1
y t+1 z t+1
y xy



where x, y, z are real numbers whose sum is 0. +z z +x x
F (t) = 1 .
Since cosh x = cosh(y + z), etc., the inequality can be [x + y + z] t+1
rewritten as
Then by the Power Mean Inequality, F (t) ≥ F (0) = 1.
(i) sinh y sinh z + sinh z sinh x + sinh x sinh y ≤ 0.
Exercise. Prove more generally that
Since (i) is obviously valid if at least one of x, y, z = 0, we can
assume that xyz 6= 0 and x, y > 0. Since z = −(x + y), (i) xt+1 y t+1 z t+1 (x + y + z)t+1
becomes csch x + csch y ≥ csch(x + y) for all x, y > 0. This + + ≥ ,
follows immediately since csch t is a decreasing function for at bt ct (a + b + c)t
all t > 0.
where x, y, z, a, b, c > 0 and t ≥ 0.
Exercise. Prove that
15. Determine the maximum value of
v w w u u v 1 1 1
+ + + + + ≥u+ +v+ +w+ S = 3(a3 + b2 + c) − 2(bc + ca + ab),
w v u w v u u v w
where u, v, w > 0 and uvw = 1. where 1 ≥ a, b, c ≥ 0.

12. It is known and elementary that in a triangle, the longest Here, S ≤ 3(a + b + c) − 2(bc + ca + ab). Since this latter
median is the one to the shortest side and the shortest median expression is linear in each of a, b, c, its maximum value is
is the one to the longest side. Determine whether or not the taken on for a, b, c = 0 or 1. Hence the maximum is 6−2 = 4.
longest median of a tetrahedron is the one to the smallest Exercise. Determine the maximum value of
area face and the shortest median is the one to the largest
area face. S = 4(a4 + b4 + c4 + d4 ) − (a2 bc + b2 cd + c2 da + d2 ab)
Let the sides of tetrahedron P ABC be given by P A = a, −(a2 b + b2 c + c2 d + d2 a),
P B = b, P C = c, P C = d, CA = e, and AB = f . The
~ ~ C|
median mp from P is given by |A+B+
~
~ B,
where A, ~ C~ are where 1 ≥ a, b, c, d ≥ 0.
3
vectors from P to A, B, C respectively. Then 16. Determine the maximum and minimum values of
~ +B
9m2p = |A ~ + C|
~ 2 = 3(a2 + b2 + c2 ) − (d2 + e2 + f 2 ) sin A + sin B + sin C + sin D + sin E + sin F,

28
 π
where A + B + C + D + E + F = 2π and 2 ≥ A, B, C, D, Summer Institute for
E, F ≥ 0.
Mathematics at the
Here we get a quick solution by applying Karamata’s In-
equality [3]. If two vectors A ~ and B~ having n components, University of
ai and bi , are arranged in non-increasing magnitude are such Washington
that
Xk Xk SIMUW is seeking applications from talented and en-
ai ≥ bi , k = 1, 2, . . . , n − 1, thusiastic high school students for its 2004 summer
i=1 i=1 program.
and Admission is competitive.
n n
X X Twenty-four students will be se-
ai = bi , lected from Washington, British
i=1 i=1 Columbia, Oregon, Idaho, and
~ majorizes B ~ and write A ~  B.
~ We then have Alaska. Room, board, and
we say that A participation in all activities are
for a convex function F (x) that completely free for all admitted
participants.
F (a1 ) + F (a2 ) + · · · + F (an ) ≥ F (b1 ) + F (b2 ) + · · · + F (bn ).
Six weeks of classroom activi-
If F (x) is concave, the inequality is reversed. ties, special lectures, and related
activities are led by mathemati-
Since sin x is concave in [0, π/2], and cians and other scientists with
the help of graduate and under-
π π π π  graduate teaching assistants.
, , , , 0, 0  (A, B, C, D, E, F ) SIMUW activities are de-
2 2 2 2
2π 2π 2π 2π 2π 2π
 Students experimenting signed to allow students to par-
 , , , , , . with boomerang. ticipate in the experience of
6 6 6 6 6 6 mathematical inquiry and to be
√ immersed in the world of math-
The maximum value is 6 sin π3 or 3 3 and the minimum value ematics. Topics are accessible yet of sufficient sophistication
is 4 sin π2 or 4. to be challenging.

Exercise. Determine the extreme values of a5 + b5 + c5 +

d5 + e5 + f 5 given that a, b, c, d, e, f , are distinct positive
integers with sum 36.
References:
1. D.S. Mitrinovic, Analytic Inequalities, Springer-Verlag, Heidel-
berg, 1970, p. 14.
2. E.F. Beckenbach and R. Bellman, Inequalities, Springer-Verlag,
New York, 1965, p. 18.
3. A.W. Marshall and I. Olkin, Inedqualities: Theory of Majoriza-
tion and Its Applications, Academic Press, New York, 1979.
2003 SIMUW participants
Murray S. Klamkin has a long and distinguished career
in both industry and academia. He is known primarily as a Students will gain a full appreciation of the nature of math-
problem solver, editing the problem corners of many journals ematics: its wide-ranging content, the intrinsic beauty of its
over the years. He put this talent to good use in leading ideas, the nature of mathematical argument and proof, and
the USA team in the IMO, chairing the USAMO committee the surprising power of mathematics within the sciences and
and authoring several books on mathematics competitions. beyond.
He is particularly fond of triangle inequalities and spherical To obtain more information and application materials, con-
geometry. (Andy Liu) tact us at:
http://www.math.washington.edu/∼simuw
SIMUW
Department of Mathematics
University of Washington
Box 354350
Seattle, WA 98195–4350
Phone: (206) 992–5469
Mother to her daughter: “Why does the tablecloth you just put Fax: (206) 543–0397
on the table have the word ‘truth’ written on it?” E-mail: simuw@math.washington.edu
Daughter: “Because I want to turn the table into a truth table!”
The 2004 SIMUW program runs from June 20th to
July 31st.

29