Putnam 2017, Problem A-3

Let \( a \) and \( b \) be real numbers with \( a < b \), and let \( f \) and \( g \) be continuous functions from \( [a,b] \) to \( (0,\infty) \) such that \( \int_a^b f(x) \, dx = \int_a^b g(x) \, dx \) but \( f \neq g \). For every positive integer \( n \), define  
\[
I_n = \int_a^b \frac{(f(x))^{n+1}}{(g(x))^n} \, dx.
\]  
Show that \( I_1, I_2, I_3, \ldots \) is an increasing sequence with \( \lim_{n \to \infty} I_n = \infty \).

Solution:
### First solution
Extend the definition of \( I_n \) to \( n = 0 \), so that \( I_0 = \int_a^b f(x) dx > 0 \). Since \( \int_a^b (f(x) - g(x)) dx = 0 \), we have  
\[
I_1 - I_0 = \int_a^b \frac{f(x)}{g(x)} (f(x) - g(x)) dx = \int_a^b \frac{(f(x) - g(x))^2}{g(x)} dx > 0,
\]  
where the inequality follows from the fact that the integrand is a nonnegative continuous function on \( [a, b] \) that is not identically \( 0 \). Now for \( n \geq 0 \), the Cauchy–Schwarz inequality gives  
\[
I_n I_{n+2} = \left( \int_a^b \frac{(f(x))^{n+1}}{(g(x))^n} dx \right) \left( \int_a^b \frac{(f(x))^{n+3}}{(g(x))^{n+2}} dx \right) \geq \left( \int_a^b \frac{(f(x))^{n+2}}{(g(x))^{n+1}} dx \right)^2 = I_{n+1}^2.
\]  

It follows that the sequence \( \{I_{n+1}/I_n\}_{n=0}^\infty \) is nondecreasing. Since \( I_1/I_0 > 1 \), this implies that \( I_{n+1} > I_n \) for all \( n \); also, \( I_n/I_0 = \prod_{k=0}^{n-1} (I_{k+1}/I_k) \geq (I_1/I_0)^n \), and so \( \lim_{n \to \infty} I_n = \infty \) since \( I_1/I_0 > 1 \) and \( I_0 > 0 \).  

### Remark
Noam Elkies suggests the following variant of the previous solution, which eliminates the need to separately check that \( I_1 > I_0 \). First, the proof that \( I_n I_{n+2} \geq I_{n+1}^2 \) applies also for \( n = -1 \) under the convention that \( I_{-1} = \int_a^b g(x) dx \) (as in the fourth solution below). Second, this equality must be strict for each \( n \geq -1 \): otherwise, the equality condition in Cauchy–Schwarz would imply that \( g(x) = c f(x) \) identically for some \( c > 0 \), and the equality \( \int_a^b f(x) dx = \int_a^b g(x) dx \) would then force \( c = 1 \), contrary to assumption. Consequently, the sequence \( I_{n+1}/I_n \) is strictly increasing; since \( I_0/I_{-1} = 1 \), it follows that for \( n \geq 0 \), we again have \( I_{n+1}/I_n \geq I_1/I_0 > 1 \) and so on.  

### Second solution
(from Art of Problem Solving, user MSTang) Since \( \int_a^b (f(x) - g(x)) dx = 0 \), we have  
\[
I_{n+1} - I_n = \int_a^b \left( \frac{(f(x))^{n+2}}{(g(x))^{n+1}} - \frac{(f(x))^{n+1}}{(g(x))^n} \right) dx = \int_a^b \frac{(f(x))^{n+1}}{(g(x))^{n+1}} (f(x) - g(x)) dx
\]  
\[
= \int_a^b \left( \frac{(f(x))^{n+1}}{(g(x))^{n+1}} - 1 \right) (f(x) - g(x)) dx = \int_a^b \frac{(f(x) - g(x))^2 ((f(x))^n + \cdots + g(x)^n)}{(g(x))^{n+1}} dx.
\]  

The integrand is continuous, nonnegative, and not identically zero; hence \( I_{n+1} - I_n > 0 \).  

To prove that \( \lim_{n \to \infty} I_n = \infty \), note that we cannot have \( f(x) \leq g(x) \) identically, as then the equality \( \int_a^b f(x) dx = \int_a^b g(x) dx \) would imply \( f(x) = g(x) \) identically. That is, there exists some \( t \in [a, b] \) such that \( f(t) > g(t) \). By continuity, there exist a quantity \( c > 1 \) and an interval \( J = [t_0, t_1] \) in \( [a, b] \) such that \( f(x) \geq c g(x) \) for all \( x \in J \). We then have  
\[
I_n \geq \int_{t_0}^{t_1} \frac{(f(x))^{n+1}}{(g(x))^n} dx \geq c^n \int_{t_0}^{t_1} f(x) dx;
\]  
since \( f(x) > 0 \) everywhere, we have \( \int_{t_0}^{t_1} f(x) dx > 0 \) and hence \( I_n \) is bounded below by a quantity which tends to \( \infty \).  

### Remark
One can also give a variation of the second half of the solution which shows directly that \( I_{n+1} - I_n \geq c^n d \) for some \( c > 1, d > 0 \), thus proving both assertions at once.  

### Third solution
(from David Savitt, via Art of Problem Solving) Extend the definition of \( I_n \) to all real \( n \), and note that  
\[
I_{-1} = \int_a^b g(x) dx = \int_a^b f(x) dx = I_0.
\]  

By writing  
\[
I_n = \int_a^b \exp((n + 1) \log f(x) - n \log g(x)) dx,
\]  
we see that the integrand is a strictly convex function of \( n \), as then is \( I_n \). It follows that \( I_n \) is strictly increasing and unbounded for \( n \geq 1 \).  

### Fourth solution
(by David Rusin) Again, extend the definition of \( I_n \) to \( n = -1 \). Now note that for \( n \geq 0 \) and \( x \in [a, b] \), we have  
\[
(f(x) - g(x)) \left( \left( \frac{f(x)}{g(x)} \right)^{n+1} - \left( \frac{f(x)}{g(x)} \right)^n \right) \geq 0
\]  
because both factors have the same sign (depending on the comparison between \( f(x) \) and \( g(x) \)); moreover, equality only occurs when \( f(x) = g(x) \). Since \( f \) and \( g \) are not identically equal, we deduce that  
\[
I_{n+1} - I_n > I_n - I_{n-1}
\]  
and so in particular  
\[
I_{n+1} - I_n \geq I_1 - I_0 > I_0 - I_{-1} = 0.
\]  

This proves both claims.  

### Remark
This problem appeared in 2005 on an undergraduate math olympiad in Brazil. See https://artofproblemsolving.com/community/c7h57686p354392 for discussion.