Some 1991 Putnam solutions, finally committed to electronic form ten years
(minus two weeks) later...  rusin@math.niu.edu

A1. First compute more generally this area:


<Please pardon our dust while we find the time to do the typing.>

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For Problem B-6 we quote this email:

Date: Mon, 17 Sep 2001 15:36:13 +0200
From: "Gerald Petit" <gerald.petit@industrie.gouv.fr>
To: jscholes@kalva.demon.co.uk
Subject: Putnam Problem: 1991 (B6)

Dear Mr Scholes,

Here is a suggestion concerning a Putnam problem (1991 B6).

Thank you in advance for possible comments.

Yours sincerely,

Gérald Petit
petitge@yahoo.com

Problem
-------

"Let a and b be positive numbers. Find the largest number c, in terms of
a and b, such that for all u with 0 < |u| <= c and for all x
with 0 < x < 1, we have: 

(a^x).(b^(1-x)) <= a sinh ux/sinh u + b sinh u(1 - x)/sinh u. "

You suggested a solution to this at:

http://www.kalva.demon.co.uk/putnam/psoln/psol9112.html

on which you commented: "I do not like this solution (which is adapted
from the official solution in the American Mathematical Journal). It
gives no real clue as to what is going on. Better motivated or more
enlightening solutions would be welcome." 

I don't know whether you will find the following more attractive:

1) Put f(x;m,n,r)=m.cosh(rx)+n.sinh(rx) for x in [0,1] and: m,n,r real
numbers, r>=0.

Put F(x)=(a^x).(b^(1-x)) 
and G(x)= a sinh ux/sinh u + b sinh u(1 - x)/sinh u

F and G belong to the family f(.;m,n,r) (obviously it's enough to
suppose u>=0 in our problem).

Moreoever members of the family f(.;m,n,r) which take the same value at
x=0 (say, A>=0) and x=1 (=say, B>=0) - which is the case in our problem
- can be written as:

f(x;A,B,r)= A.cosh(rx)+((B-(A.coshr))/sinhr).sinhrx

Now it is requested to find a condition on u such that:

F(x)<=G(x) for all x in [0,1]

With:

F(0)=G(0)=b and F(1)=G(1)=a in our particular problem. We can suppose
b>=a.

1) We first prove that: 

if f(x;A,B,r)<=g(x;A,B,s) for all x in [0,1], then r>=s.

For this since f(0;r)=g(0;s) and f(1;r)=g(1;s), we have f'(0;r)<=f'(0;s)
and f'(1;r)>=f'(1;s), that is:

((B-(A.coshr))/sinhr).rcoshr <= ((B-(A.coshs))/sinhs).scoshs and:

rA.sinhr+ ((B-(A.coshr))/sinhr).rcoshr >= sA.sinhs+
((B-(A.coshs))/sinhs).scoshs

Thus rA.sinhr >= sA.sinhs, hence r>=s since r -> r.sinhr increases.

Applying this to our problem where r=Log(b/a) and s=u we find the
necessary condition u<=Log(b/a), for F(x)<=G(x) to be true for all x in
[0,1].

2) In order to prove the converse inequality we next use a little lemma:

Let f and g be differentiable and convex functions on [0,1] such that:

f(1)=g(1) <= f(0)=g(0) and f'(0)<=g'(0), 

and suppose moreoever that f and g are both decreasing.

Then f(x)<=g(x) for all x in [0,1].

Proof of lemma : 
---------------

When making a drawing this looks pretty clear; I have not found a very
straightforward proof of this, though.

a) Look first at U(x)= f(x)-(g'(0).x)-g(0). 

Since U'(x)=f'(x)-g'(0) increases, either f'(1)<=g'(0) in which case U
decresases, or f'(1)>g'(0) and U reaches a minimum. 

In the first case since U(0)=0, we would have: 

U(1)= f(1)-g'(0)-g(0)=g(1)-g(0)-g'(0)<U(0)=0, which is impossible
because g is convex. Thus U reaches a minimum and then increases up to
U(1)= f(1)-g'(0)-g(0)=g(1)-g'(0)-g(0)>=0.

Therefore, there is some x_0, verifying U(x_0)=0 and such that:

For all x in [0, x_0], U(x)<=0, that is: 

f(x)<=(g'(0).x)-g(0)<=g(x) since g is convex.

b) Look next at V(x)= f(x)-g'(1).(x-1)-g(1). 

- Suppose f'(1)<g'(1). Then V is decreasing from: 

V(0)=f(0)+g'(1)-g(1)=g(0)+g'(1)-g(1)>=0 (g is convex) to V(1)=0.

Thus there is some x_1 verifying V(x_1)=0 such that:

For all x in [x_1, 1], V(x)<=0, that is: 

f(x)<=g'(1).(x-1)-g(1)<=g(x) since g is convex.

- Suppose f'(1)>=g'(1). Then V reaches a minimum and there is also some
x_1 verifying V(x_1)=0 and such that:

For all x in [x_1, 1], V(x)<=0.

c) Finally compute f(x_1)-f(x_0).

f(x_1)-f(x_0)= g'(1).(x_1-1) - g(1) - g'(0).x_0 + g(0);
f(x_1)-f(x_0) >= -g(1) - g'(0).x_0 + g(0) >= -g(1) - g'(0) + g(0);
-g(1) - g'(0) + g(0) >= -g(1) - (g(1)-g(0)) + g(0) = 2 (g(0)-g(1))>=0.

Thus f(x_1)-f(x_0)>=0 and since f decreases, x_1<=x_0.

d) Finally:

For all x in [0, x_1] f(x)<=g(x) because x_1<=x_0.
And for all x in [x_1, 1], we also have f(x)<=g(x) because V(x)<=0.

QED.

NOTE: we can reduce the hypotheses, since we only used (i) g convex (ii) 
f convex and decreasing (iii) f(1)<=g(1)<= f(0)=g(0), f'(0)<=g'(0) (iv)
g'(1)<=0.

3) Now suppose that 0<=u<=Log(b/a). 

- F and G are convex and F is decreasing;

- F(0)=G(0)=b >= F(1)=G(1)=a;

- F'(0)= b.Log(a/b), G'(0)=(u/sinhu).(a-b.coshu)

Put	V(u)=b.Log(a/b).sinhu - u.(a-b.coshu) then:
	V'(u)= b.(Log(a/b)+1).coshu - a + u.b.sinhu

Since a<=b, b.(Log(a/b)+1)>=a thus V'(u)>=0 and V increases. 
V(u)<=V(Log(b/a))= 0; Thus F'(0) <= G'(0).

- G'(1)= (u/sinhu).(a.coshu-b)
Since u<=Log(b/a), a.coshu-b <= (1/2).(a^2-b^2)<=0. Thus G'(1)<=0.

Using the preceding lemma we conclude that:

For all x in [0,1], F(x)<=G(x).

QED.