Advanced book on Mathematics Olympiad

600 Real Analysis

which implies thatA=B=0, and thereforey 1 =y 2 , as desired.
(M. Ghermanescu, ̆ Ecua ̧tii Diferen ̧tiale(Differential Equations), Editura Didactic ̆a
̧si Pedagogica, Bucharest, 1963) ̆

568.LetF(t)=

∫t
0 f(s)dsbe the antiderivative offthat is 0 at the origin. The inequality
from the problem can be written as

F′(t)

√

1 + 2 F(t)

≤ 1 ,

which now reminds us of the method of separation of variables. The left-hand side is
the derivative of

√

1 + 2 F(t), a function whose value at the origin is 1. Its derivative is
dominated by the derivative ofg(t)=t+1, another function whose value at the origin
is also 1. Integrating, we obtain
√
1 + 2 F(t)≤t+ 1.

Look at the relation from the statement. It says thatf(t)≤

√

1 + 2 F(t). Hence the
conclusion.
(P.N. de Souza, J.N. Silva,Berkeley Problems in Mathematics, Springer, 2004)

569.We will use the “integrating factor’’ex. The inequalityf′′(x)ex+ 2 f′(x)ex+
f(x)ex≥0 is equivalent to(f (x)ex)′′≥0. So the functionf(x)exis convex, which
means that it attains its maximum at one of the endpoints of the interval of definition.
We therefore havef(x)ex≤max(f ( 0 ), f ( 1 )e)=0, and sof(x)≤0 for allx∈[ 0 , 1 ].
(P.N. de Souza, J.N. Silva,Berkeley Problems in Mathematics, Springer, 2004)

570.Assume that such a function exists. Becausef′(x)=f(f(x)) >0, the function is
strictly increasing.
The monotonicity and the positivity offimply thatf(f(x)) > f( 0 )for allx. Thus
f( 0 )is a lower bound forf′(x). Integrating the inequalityf( 0 )<f′(x)forx<0, we
obtain

f(x)<f( 0 )+f( 0 )x=(x+ 1 )f ( 0 ).

But then forx ≤−1, we would havef(x)≤ 0, contradicting the hypothesis that
f(x) >0 for allx. We conclude that such a function does not exist.
(9th International Mathematics Competition for University Students, 2002)

571.We use the separation of variables, writing the relation from the statement as

∑n

i= 1

P′(x)

P(x)−xi

=

n^2

x

.

Integrating, we obtain