Targil 12 – some inequalities.
1. a) What is greater: eπ or πe ?
b) The sum of several natural numbers is 2008.
What is largest possible value of their product?
a) First solution. ex is a strictly convex function, so it is above the tangent
line at 1. Tangent line at 1 is ex. Hence ex ≥ ex , and since convexity is strict
the equality may happen only when x = 1.
Hence eπ/e > e·π/e = π .
So eπ > πe.
Second solution. The question is, actually, what is greater, e e or   .
To find compute the derivative of x x function (or, if you don’t like the hard
work, the derivative of it logarithm, x ln x ). You will see that the derivative
is positive before e, negative after e, and 0 at e.
So e e is the greatest of all x x .
b) If k>4, then k(k–2) is bigger then k, so there should be no numbers bigger
than 4. If k = 4, we can replace it by 2·2 and still have the same product.
Hence WLOG, we can assume we have only 2’s and 3’s. But 3·3 is greater
than 2·2·2, so we have less than three 2’s.
2004 is divisible by 3, so we can have two 2’s and 668 times 3, or one 4 and
668 times 3. The product is 3668·4 .
Remark. The question is, morally, which numbers, k’s or m’s, are more
useful, which is bigger, mk or km, ore in other words, k k or m m . The
answer is, that the best number is e, but as long as we are bound to use
integer numbers, the best is 3, the next two are 2 and 4, which are equivalent
since 2 2  4 4 and other number are worse since they are worse
approximations of e. If the question would be about real, and not integer
numbers, we would get equal numbers (by AM-GM) very close to e.
2. Let 0 < x < π/2. What is greater: tan(sin(x)) or sin(tan(x)) ?
First solution. Let us start with a guess. On the domain sin goes upwards
from 0 to 1 so tan(sin) goes monotonically upwards. While tan(x) goes
upwards from 0 to infinity hence sin(tan) oscillates on the domain, dives to
negative and returns to positive many times. So, at least sometimes the first
function is bigger, and there wouldn’t be a point in this question unless it is
always bigger. So, we shall prove that tan(sin(x)) > sin(tan(x)).
It is enough to prove the claim on the domain where both functions are still
growing, i.e. when tan(x) ≤ π/4. At arctan(π/4) the sin(tan(x)) reaches its
global maximum, 1, while tan(sin(x)) keeps growing.
Denote y = tan(sin(x)). Then sin(x) = arctan(y) (*).
Hence the main claim is arcsin(y) > tan(x).
It is enough to prove that the rate of growing of LHS is higher, i. e.
dx
dy

2
cos x
1  y2
cos 2 x dy
1
1  y 2 dx
While from (*) we have
cos x  dx 
Hence we can substitute
dy
1  y2
dy
into the claim:
dx
1  y2
3
1  cos x
1  y2
But from (*) we can express cos x  1  arctan 2 y so now we have to prove
arrive to an innocently looking inequality, with no composed functions:
2
3 2 1 y
2
1  1  arctan y 
1  y2
Where 0 < y < 1. Here we have square roots twice, so let’s take the square:
3
1  y2
 1  arctan 2 y 
2
1  y 2 
1
y2
1
But we can write
1  y2
1  y2
Hence if z = arctan y, then we can rewrite the claim as
1  tan z 1  sin z 1  sin z   1  z 
2
2
2
2
 cos 2 z  sin 2 z 
2
2 3

  cos z   1  z 
2
cos z


2 3
cos  2 z  cos 2 z  1  z 2 
3
3
cos  2 z   cos2  2 z 
1  cos  2 z 
 cos  2 z 
 1  z 2 
2
2
3
cos  4 z   2cos  2 z   1
 1  z 2 
4
Here 0 < z < arctan(1) = 1/ 2 < 1.
So, we shall estimate the left hand side by its Taylor series. We know from
Lagrange remainder that if we take 4k+1 terms we shall get upper bound for
cosine. Hence:
2
4
6
8
cos  4 z   2cos  2 z   1 1 
4z   4z   4z   4z  

 1  1 




4
4 
2
24
6!
8! 
2
4
6
8
1  2z  2z  2z  2z  
 1 




2 
2
24
6!
8! 
43  23 4 45  25 6 4 7  2 7 8
22 6 43 8
2
 1  3z 
z 
z 
z  1  3z 2  3z 4 
z 
z
24
6!
8!
15
105
So, to achieve happiness, i. e. to prove that LHS < (1 – z2)3 it is enough to
prove that
22 6 43 8
1  3z 2  3z 4 
z 
z  1  3z 2  3z 4  z 6
15
105
43 8 7 6
z  z
105
15
735
z2 
645
It is true in the domain which we consider. QED.
Sorry people, less Taylor terms are simply not enough. And yes, I did the
computations by hands.
Second solution (the official one). Let f(x) = tan(sin(x)) – sin(tan(x)). Then
cos  tan  x   cos3 x  cos 2  sin x  cos  tan  x  
cos x
f '( x) 


cos 2  sin x 
cos 2 x
cos 2  sin x  cos 2 x
Let 0 < x < arctan(π/2). Since cos is concave (sad) on (0 , π/2) we get
cos  tan x   2cos  sin x 
tan x  2sin x
3 cos 2  sin x  cos tan  x  
 cos
 cos  x 


3
3
tan x  2sin x
 x , which is because
3
/
1
 tan x  2sin x  1  1

  2  2cos x   3
 cos x  cos x  1


3
cos 2 x
 3  cos x

The last inequality follows from
3. We are given 5 positive real numbers: a, b, c, d, e such that
a2 + b2 + c2 = d2 + e2
a4 + b4 + c4 = d4 + e4
What is greater: a3 + b3 + c3 or d3 + e3 ?
First solution. Let A=a2, B=b2, C=c2, D=d2, E=e2.
P=A+B+C = D+E , Q = A2+B2+C2 = D2 + E2.
Consider in the {(x, y, z)} space the set defined by
x+y+z=P
x2 + y2 + z2 = Q
The first equation is a plane, the second is a sphere. So, the intersection is a
circle. It cuts coordinate planes at points (D, E, 0), (E, D, 0), (D, 0, E), etc…
Hence it has three symmetric arcs in the positive domain x, y, z > 0 (one arc
is where x is the greatest; another where y is greatest; the third is where z is
greatest). So, let us assume that A > B > C and D > E so that points
(A, B, C) and (D, E, 0) will be one the same arc of the circle (actually, on the
same half of the same arc, because y > z in both cases).
Let (X(t) , Y(t) , Z(t)) be a parametric curve, going along that arc from the
point (D, E, 0) to the point (A, B, C), which means, for instance, that
(X(0) , Y(0) , Z(0)) = (D, E, 0)
(X(1) , Y(1) , Z(1)) = (A, B, C)
We shall also assume that the curve goes always forward along the arc with
constant speed v. By upper dot we denote, as usual, derivative over t.
So, since X + Y + Z and X2 + Y2 + Z2 remain constant during the motion
X Y  Z  0
2 XX  2YY  2ZZ  0
So the vector ( X , Y , Z ) is orthogonal to both (1, 1, 1) and (X, Y, Z) and so it
is proportional to their vector product (Y  Z , Z  X , X  Y ) .
But we know that Z is growing along the way, and X is the greatest of 3, so
the coefficient of proportion is positive.
So, we have computed ( X , Y , Z ) up to a positive coefficient.
The question which really bothers us is whether the value of the function
F(t) = X 3/2 + Y 3/2 + Z 3/2
is greater at the beginning or at the end of the motion?
Let us derive it and find its sign.
3
F (t )  X X  Y Y  Z Z  a Y  Z  X   Z  X  Y   X  Y  Z
2
Here a is a positive coefficient so it doesn’t influence the sign.
Denote x  X , y  Y , z  Z . Then, since X > Y > Z > 0 on our half-arc,
so also x > y > z > 0. Then
Y  Z  X   Z  X  Y   X  Y  Z   y 2  z 2  x   z 2  x 2  y   x 2  y 2  z 

 
 y 2 x  x 2 y  z 2  x  y    x  y  x  y  z   x  y    xy  z 2   x  y  z  
   x  y  x  z  y  z   0
(Of course, this expression was Vandermonde of size 3).
So, F (t )  0 , and the value if
d3 + e3 = F(0) > F(1) = a3 + b3 + c3.
Second solution (the official). WLOG, we may assume that a≥b≥c and d≥e.
Let c2 = e2 + ∆. Then d2 = a2 + b2 + ∆.
Hence a 4  b 4   e 2      a 2  b 2     e 4
4
2
2e2   2a 2b 2  2  a 2  b 2  
a 2b2
 2
a  b2  e2
2
1
1
But a 2  b2  e2   a 2  b2  c 2    d 2  e2    d 2  e2   0
3
2
6
So ∆ is negative.
2
2
2
2
a 2b 2
a 2e2  b 2e2  e4  a 2b 2  a  e  e  b 
2
2
0c e  2


a  b2  e2
a 2  b2  e2
a 2  b2  e2
So a > e > b.
a 2b2
a 2b2
2
2
2
2
2
Therefore d  a  b  2
 a  b  2  a2 .
2
2
a b e
a
Hence a > d ≥ e > b ≥ c .
Consider the function
f(x) = ax + bx + cx – dx – ex
WLOG a = 1(if not, divide everything by ax).
We shall prove that this function has only 2 zeroes on the real line, 2 and 4,
and that f changes its sign at those points. Suppose the contrary. Then, by

Rolle’s theorem, there f ' has at least two distinct zeroes, x1 < x2 ,
f '  x1   f '  x2   0 .
f '  x   ln b  b x  ln c  c x  ln d  d x  ln e  e x
Hence ln b  b xi  ln c  c xi  ln d  d xi  ln e  e xi for i = 1, 2.
But 1 > d ≥ e > b ≥ c . So
  ln b  b x1    ln c  c x1  b x1  x2  e x1  x2    ln d  d x1    ln e  e x1 ,
  ln b  b x2    ln c  c x2
  ln d  d x2    ln e  e x2
a contradiction.
Hence f(x) has constant sign at intervals  ,2  ,  2,4  ,  4,   . It is positive
at 0, so it is negative at 3.
4. a) Let {xi} be a decreasing sequence of positive numbers.
n
n
xi
2
x

Prove that


i
i
i 1
i 1
b) Prove that there exists a universal constant C, such that for each
decreasing sequence of positive numbers


n
1
2
xi

 xi  C 
m i m
m 1
i 1
2
 n xi 
Solution. a) Take the square:  x    
i 1
 i 1 i 
But if You replace all numbers by lower numbers with higher indices:
n
2
i
2
x2 
1
x
x 
 n xi 
 x 
2
x2   3  2 x1  2 2  3   ... 
 2 x1 
    x1 
2
2 
3
2
3
 i 1 i 
x22 
1  x32 
1
1  x42 
1
1
1 
2
 x1 

2

2

2  2

2  2
  ...
2
2
3
2
3
4
2
3
4
It remains to estimate sum of inverse roots. It is easy to guess (for example,
if you approximate series by integral) that it is approximately n .
To make precise statement consider
n  n 1 

n  n 1

n  n 1

1 
 1


 4
n 1 
n  n 1  n
1
n  n 1
(Because always 1   1  1  4  A  B .
A B  A B
4 AB
Since 4AB   A  B  , which is same as 0   A  B  .)
Sum up such inequalities from 2 to m and You get
1 1
2
2
2
1 
m 1  


 ... 


4 1
2
3
m 1
m
2
2
2
2
1
So m 


 ... 

1
2
3
m 1
m
(The last formula could be obtained with less subtle technique, but technique
is what we want to learn). Now we can finish the proof:
2
2
2
x22 
1  x32 
1
1 
 n xi 
2
2
2
2
2

x

2


2

2


1



  ...  x1  x2  x3  x4 ...


2
2
3
2
3
 i 1 i 
b)




i
1
1 
xi
1
1
2
x


x






i
i
m i m
m i m i  1  m i 1 m1 m i  1  m
m 1
m 1
All we have to do is to estimate
1
i
1
1
1 i 1
1
dx

 

~

i  1  m i m1 m
1 m 0 x 1 x
m 1 m
1 
i
i i
Because it is approximately Riemann’s sum for that integral. Actually, if we
divide the interval [0,1] into i equal parts, then in the part number m we have
m
1 m
1
1
1
and


x  ,1  x  1   hence
i
i i
m
1 m
x 1 x
1 
i
i i
1
i
1
1
1 i 1
1
dx

 



i  1  m i m1 m
1 m 0 x 1 x
m 1 m
1 
i
i i
So, it remains to compute that integral, and it will serve as (a pretty tight)
universal constant. It is done by a trigo substitution: x = sin2t
dx = 2sint cost dt
1

0

dx

x 1 x
2
 2dt   .
0
There is a more elementary way to find a less tight bound:
i
i/2
1
1
1
1
1 i/2 1
1


2


2
2
2 i/2  4



m
i

1

m
m
i

1

m
i
/
2
m
i
/
2
m1
m1
m1

5**. Let {an} be a sequence of positive numbers, such that
a
n 1

a) Prove that

n 1
n
 .

n
a1  a2  ...  an  e an .
n 1
b) Prove that the constant in the inequality can’t be improved, i. e. for any
positive  we can find a sequence of positive numbers s.t.

a
n 1


n 1
n
  and

n
a1  a2  ...  an   e     an
n 1
This inequality is called Carleman inequality. Carleman is a famous
Swedish mathematician.
This inequality was given at IMC and it was the most expensive problem in
its year. We shall start with b) part, since it is easier and gives a clue to a).
1
where u is only slightly bigger then 1.
nu
Then the sum of an is bounded, but very big, so the first few terms are only a
small part of the sum. For large n’s,
u
u
1
e




u
n a  a  ...  a 
1
2
n
n
 ~  n   e an ~ ean
 1  2  ...  n   
Hence for u sufficiently close to 1, the LHS and RHS of Carleman inequality
are approximately equal. QED.
b) Consider the sequence an 
This gives a clue to a part. It is very natural to try and decompose Carleman
inequality into sum of Cauchy (AM-GM) inequalities, say
b a  bn ,2 a2  ...  bn ,n an
n a  a  ...  a  n ,1 1
n b b
n ,1
n ,2  ...  bn ,n
1
2
n
n
Then, if you choose the coefficients smartly, so that
n
bn ,1  bn ,2  ...  bn ,n is less
than e times sum of all coefficients before each ak, you will get Carleman.
But how to guess those coefficients?
It is well known that Cauchy inequality becomes equality only if all the
numbers in it are the same. Assume (though it is literally wrong, but morally
almost true), that Carleman inequality becomes equality for the sequence
1
an  . So, the coefficients should be chosen so, that the numbers will
n
become equal, so an should be multiplied by n before applying Cauchy.
That was the main philosophical idea, now we shall execute it.
a) We shall start with reminding classical calculus lemmas:
Lemma 0. (1+a)n ≥ 1 + na if 1+a > 0 and n is a natural number.
(Bernoulli inequality). Equality holds only if a = 0 or n = 1.
n
n
 1   n 1
Lemma 1. bn  1    
  e.
 n  n 
n 1 n
Lemma 2.
 n! .
e
Proof of lemma 0. By induction, for n = 1 it is obvious, for n+1, supposing
it was proven for n, it follows:
(1+a)n+1 = (1+a)n(1+a) ≥ (1 + na)(1+a) = 1 + (n+1)a + na2 ≥ 1 + (n+1)a
Proof of lemma 1. It is well known that bn  e . So, it would be enough to
prove that bn is monotonically increasing, i. e. bn < bn+1.
n
n 1
 n 1  n  2 

 

 n   n 1 
 n  n  2 
n


 n  1   n  12 
n 1

1
1 
1
 1 

n  1   n  12 
n 1
That is a direct consequence of Bernoulli inequality (lemma 0).
Proof of lemma 2. Since b1  b2  ...  bn  e we get
n  1 n  n  1

 n b1  b2  ...  bn  e
n
n!
n!
n
Finally, let’s prove Carleman inequality. By Cauchy and lemma 2:
n 1 n
1  a1  2  a2  3  a3  ...  n  an
a1  a2  ...  an  n n! a1  a2  ...  an 
e
n
1  a  2  a2  3  a3  ...  n  an
Hence n a1  a2  ...  an  e 1
.
n  n  1
Sum all these inequalities. You get







k
1 
1
1
n a  a  ...  a  e
ak 
 e kak   


1
2
n
  e kak  e ak
n  1  k 1
k
n 1
k 1
nk n  n  1
k 1
n k  n
k 1
QED.